Nice problem! Let the hexagon be $ A_{1}A_{2}A_{3}A_{4}A_{5}A_{6}$. Note that $ P_{n}$ is convex (this is easy with induction). On each side $ A_{i}A_{i+1}$, where $ A_{i}=A_{i+6}$, there must exist two points, say $ X_{i}$ and $ Y_{i}$, where $ X_{i}$ is between $ Y_{i}$ and $ A_{i}$, that are vertices of $ P_{n}$. By the convexity of $ P_{n}$, $ Y_{i}X_{i+1}$ is inside $ P_{n}$ for $ i=1,2,\ldots,6$. Furthermore, $ X_{i}$ and $ Y_{i}$ are adjacent vertices, so we have $ P_{n}>[X_{1}Y_{1}X_{2}Y_{2}\ldots X_{6}Y_{6}]$. Let $ A_{i}A_{i+2}\cap A_{i+1}A_{i+3}=B_{i}$. It follows from the convexity of $ \Delta A_{i}A_{i+1}A_{i+2}$ that $ B_{1}B_{2}B_{3}B_{4}B_{5}B_{6}$ is completely contained within $ X_{1}Y_{1}\ldots X_{6}Y_{6}$, so it is contained in $ P_{n}$. But it's a straightforward calculation to find that $ [B_{1}B_{2}B_{3}B_{4}B_{5}B_{6}]=\dfrac{1}{3}$, so we're done. It seems like this bound can be improved significantly...I'm thinking it's true for 1/2.

Sorry about the 5.5 year revive darn. Also, I pleased the above poster by getting a bound of 1/2. For a $n$-gon $A_1A_2 \ldots A_n$ define its outsum to be $\sum_{i = 1}^n[A_{i}A_{i+1}A_{i+2}]$ (indices $\pmod{n}$ ofc). WLOG we clip the triangle $A_{n-1}A_nA_1$, and our polygon becomes $A_1A_2 \ldots A_{n-1}A_n'A_{n+1}$. What the the outsum of our new polygon? Well, we add $\frac{1}{4}[A_{n-1}A_nA_1]$ to our total clipped area. But note that the outsum decreases by at least $\frac{1}{2}[A_{n-1}A_nA_1]$. This is because the areas $[A_{n-2}A_{n-1}A_n], [A_nA_1A_2]$ decrease to $[A_{n-2}A_{n-1}A_n'], [A_n'A_1A_2]$ (to be exact, half as much area), while $[A_{n-1}A_nA_1]$ is replaced $[A_{n-1}A_n'A_{n+1}]+[A_n'A_{n+1}A_1] = \frac{1}{2}[A_{n-1}A_nA_{n+1}]$. But the outsum is at least positive, and the clipped area decreases by half as much as the outsum, so its at most 1/2. $%Error. "blackbox" is a bad command. $

Here is another solution obtaining a half, copy pasted from an email (hence no LaTeX). Define one cut X to be a child of another Y if one of the endpoints on X lies on the cut formed by Y. Note that at most 6 cuts are not children because of cutting the vertices/corners off. Now, the crucial lemma is that a cut's child cuts off at most 1/4 of the area the parent cuts off and this is really not hard to prove (I don't have a diagram, but basically if the parent cuts XZ where vertex Y is between, and you have YZ>ZA with A on YZ the new cut cuts off the image of XA upon a homothety with scale factor 1/2 about Z). Now, every guy's children actually sum to at most 1/2 his area. Once again this is trivial, because the heights are bounded by things like d(A,XZ)<=YZ. So create a directed graph, edges directed from parents to children. Delete edges to remove all cycles. Now we have a binary tree. The total area clipped off is at most 6*(1/24)*(1+1/2+1/4+...)=(1/4)*2=1/2 with equality never, so 1/2 is the answer.

Problem 3 of the 2010 Cono Sur Olympiad (http://www.obm.org.br/export/sites/default/como_se_preparar/provas/Cone_sul/docs/Dia1-portugues.pdf ) asks for a bound of $2/3$. This can be achieved by noting that the centroids of triangles $A_i A_{i+1} A_{i+2}$ are never removed, as well as some point in each of the sides. The convex hull of the 12 points has area at least 2/3.

Another way to get $\frac{2}{3}$ is to note that $f_n = 4\sqrt{3} \cdot \text{Area of } P_n - \sum_{l \text{ is a side of }P_n} l^2$ increases over time. This is because when we clip two sides with lengths $a, b$ and angle $\theta$ in $P_n$ to get $P_{n+1}$, the area drops by $\frac{ab \sin{\theta}}{8}$ while the sum of squared lengths drops by $a^2 + b^2 - \frac{a^2}{4} - \frac{b^2}{4} - \frac{a^2+b^2 - 2ab\cos{\theta}}{4} = \frac{a^2+b^2+ab \cos{\theta}}{2}$. The preceding claim then reduces to $ab(-\cos{\theta} + \sqrt{3} \sin{\theta}) \leq a^2 + b^2$, which is true. Note that equality holds here if and only if $a = b$ and $\theta = \frac{2\pi}{3}$, corresponding to the initial situation -- in fact, equality can hold for the first 6 clip operations, after which the angle becomes larger and this estimate is no longer tight. Going back to the original problem, suppose initially each side length is $1$, then the area of the regular hexagon is $\frac{3\sqrt{3}}{2}$. Thus the initial value of the semi-invariant is $12$, which means we have $\text{Area of } P_n \geq \sqrt{3} \geq \frac{2}{3} = \text{Area of } P_6$ for all $n$.

Let the vertices be $A_1$, $A_2$...$A_6$. Connect each $A_i$ to $A_{i+2}$, where indices are taken modulo six. We get the following diagram. [asy][asy] size(6cm, 6cm); pair A=(0,2), B=(sqrt(3),1), C=(sqrt(3),-1), D=(0,-2), E=(-sqrt(3),-1), F=(-sqrt(3),1); draw (A--B--C--D--E--F--cycle, cyan+linewidth(2)); draw (A--C--E--cycle, heavycyan+dashed+linewidth(2)); draw (B--D--F--cycle, heavycyan+dashed+linewidth(2)); dot (A, red); dot (B, red); dot (C, red); dot (D, red); dot (E, red); dot (F, red); [/asy][/asy] It is not hard to see that no cut can penetrate the interior of the inner hexagon. One can compute that the inner hexagon is $\frac{1}{3}$ of the whole hexagon. Hence, the area of $P_n$ will always be greater than $\frac{1}{3}$.

Here's just the proof of $1/3$ (from Twitch Solves ISL). Call the original hexagon $ABCDEF$. We show the area common to triangles $ACE$ and $BDF$ is in every $\mathcal{P}_n$; this solves the problem since the area is $1/3$. For every side of a clipped polygon, we define its foundation recursively as follows: $AB$, $BC$, $CD$, $DE$, $EF$, $FA$ are each their own foundation (we also call these original sides). When a new clipped edge is added, its foundation is the union of the foundations of the two edges it touches. Hence, any foundations are nonempty subsets of original sides. Claim: All foundations are in fact at most two-element sets of adjacent original sides. Proof. It's immediate by induction that any two adjacent sides have at most two elements in the union of their foundations, and if there are two, they are two adjacent original sides. $\blacksquare$ Now, if a side has foundation contained in $\{AB,BC\}$, say, then the side should be contained within triangle $ABC$. Hence the side does not touch $AC$. This proves the problem.

(https://artofproblemsolving.com/community/q1h2722725p23691094) here's another solution for $\frac{1}{2}:$ Scale up the figure of the hexagon such that it has a side length of $1.$ The hexagon now has an area of $\frac{3}{2}\sqrt{3}.$ Notice that every time $P_n$ is clipped, no side of the figure completely disappears, so the figure must touch all sides of the original regular hexagon. In addition, since the figure always remains convex, we can take the intersections of the figure and each side of the original regular hexagon, and the polygon formed will have a lesser area than the clipped figure. It suffices to find the minimum area of a hexagon that has one vertex on each side, which can be done by finding the maximum area of the six triangles that are formed by the area outside the inner hexagon but inside the outer hexagon. This is clearly equal to the maximum of $\frac{1}{2}\sin{120^{\circ}}(a(1-b)+b(1-c)+c(1-d)+e(1-f)+f(1-a))=\frac{3\sqrt{3}}{4}(a+b+c+d+e+f-ab-bc-cd-de-ef-fa)=\frac{3\sqrt{3}}{4}\left( \frac{1}{4}-\left(a-\frac{1}{2}\right)\left(b-\frac{1}{2}\right) + \frac{1}{4}-\left(b-\frac{1}{2}\right)\left(c-\frac{1}{2}\right) + \frac{1}{4}-\left(c-\frac{1}{2}\right)\left(d-\frac{1}{2}\right) + \frac{1}{4}-\left(d-\frac{1}{2}\right)\left(e-\frac{1}{2}\right) + \frac{1}{4}-\left(e-\frac{1}{2}\right)\left(f-\frac{1}{2}\right) + \frac{1}{4}-\left(f-\frac{1}{2}\right)\left(a-\frac{1}{2}\right) \right).$ Clearly we want to minimize $\sum_{cyc}\left(a-\frac{1}{2}\right)\left(b-\frac{1}{2}\right).$ We can minimize each term by setting one factor equal to $1-\frac{1}{2}=\frac{1}{2},$ and the other factor equal to $0-\frac{1}{2}=-\frac{1}{2}.$ This is achievable with $(0,1,0,1,0,1),$ which corresponds to three pairs of concurrent points at the vertices of the hexagon, which form a triangle. This triangle clearly has half the area of the hexagon, and scaling back down to a hexagon with area $1,$ the figure $P_n$ cannot have a lesser area than the triangle, which has area $\frac{1}{2}.$

A slightly different solution: Let $A_1A_2 \cdots A_6 = \mathcal P_6$ be the regular hexagon. For any $1 \le i \le 6$, define $\mathcal R_i$ as the region of triangle $A_{i-1}A_i A_{i+1}$. Claim: Fix any $j \ge 6$. Then any three vertices of $\mathcal P_j$ must be contained in some $\mathcal R_i$ and conversely, for any $\mathcal R_i$ ($1 \le i \le 6$), some three consecutive vertices of $\mathcal P_j$ will be contained in $\mathcal R_i$. Proof: Of course, the assertion is true for $\mathcal P_6$. Now if we assume the assertion is true for $\mathcal P_k$, then it is not hard to see that it is true for $\mathcal P_{k+1}$. $\square$ Now fix any $j \ge 6$. Let $P$ be the regular hexagon formed by lines $P_{i-1} P_{i+1}$ for $1 \le i \le 6$. Our claim also gives us that any side of $\mathcal P_j$ has to intersect with the interior of $P$, and it cannot be completely contained in some side of $P$. Moreover, each of the regions $\mathcal R_i$ contains at least one vertex of $\mathcal P_j$. Now I am pretty sure that this is enough to imply that area of $\mathcal P_j$ is more than the area of $P$ (though, I can't get how to prove this rigorously). Now it is not hard to prove that area of $P$ is $\frac{1}{3}$, so we are done. $\blacksquare$

Let the hexagon be $ABCDEF$. Consider the hexagon in the intersection of $\triangle ACE$ and $\triangle BDF$, which has exactly $\frac13$ of the desired area. We show that it lies in all $\mathcal{P}_n$. Now, consider the vertices of the hexagon as a cyclic set of vectors. Clipping is equivalent to replacing adjacent vectors $a, b, c$ with $a, \frac{a+b}{2}, \frac{b+c}{2}, c$. Claim: Suppose we have adjacent vectors $a, b, c, d$. Then at no point does clipping give us a vector which is a weighted average containing both $a$ and $d$. Proof. There are always two vectors not containing either components of $a$ or $d$ between the two under the operation of clipping. $\blacksquare$ Noteably, if $v$ and $u$ are two adjacent vertices in $\mathcal{P_n}$, then the subset of $\{a, b, \dots, f\}$ contained in one of them has at most three elements. As such, all edges of $\mathcal{P}_n$ lie in some triangle formed by three adjacent vertices of the original hexagon.