Notice that the polygon has perimeter at most $4$ as if one convex polygon lies in another it has smaller perimeter. Let the sides of the polygon be $x_1,x_2,\dots,x_n$. Since $(x_1+x_2)+(x_2+x_3)+\dots+(x_n+x_1)\leq 8$ there must exist a index $i$ for which $x_i+x_{i+1} \leq 8/n$. Now we can just choose the triangle with side lengths $x_i$ and $x_{i+1}$ that has area $x_ix_{i+1}\sin(\theta)/2\leq 8/n^2$.

I will show that there always is a triangle with area $< \frac{16\pi}{n^3}$. Assume not, let the vertices of the polygon be $A_i$, and consider $l_i$ to be the length of the side connecting $A_i$ and $A_{i+1}$.Let $\alpha_i=\angle A_{i-1}A_iA_{i+1}$ Considering a triangle formed by 3 adjacent vertices, it's area is equal to $\frac{l_{i-1}l_{i}\sin{\alpha_i}}{2}$. Taking the product we get $\prod \frac{l_{i-1}l_{i}\sin{\alpha_i}}{2} \ge (\frac{16\pi}{n^3})^n$, or regrouping that $\prod{l_i}^2\prod{\sin{\alpha_i}} \ge (\frac{32\pi}{n^3})^n$, now knowing that $\alpha_i \le \pi$ we get $\sin{\alpha_i}=\sin{(\pi-\alpha_i)} \le \pi-\alpha_i$, thus $\prod{\sin{\alpha_i}} \le \prod{(\pi-\alpha_i)}$, and by $AM-GM$ that $\prod{(\pi-\alpha_i)} \le (\frac{\sum_{i=1}^{n} {(\pi-\alpha_i})}{n})^n=(\frac{2\pi}{n})^n$, and substituting this back we get that $\prod {l_i} \ge (\frac{4}{n})^n$, but by $AM-GM$ and $\sum{l_i} \le 4$ we get that $\prod {l_i} \le (\frac{4}{n})^n$ however equality, can't hold everywhere. This is enough to show the inequality for $n \ge 7$, and for other $n$ just use $\prod{\sin{\alpha_i}} \le \prod{\sin{\frac{2\pi}{n}}}$, and now instead of the $\sin{x}<x$ idea just substitute the actual values. Looks like asymptotically, $n^3$ is the best power we can get, however the constant seems to not be the best one, as equality never holds.