Let the sides of one face be $ a$, $ b$ and $ A$ be the area of that face. Thus $ A=ab\sin(\theta)\le ab$. Let $ h$ be the distance from this face to the opposite face. Then we have $ Ah=216$. Project this face onto the plane of the opposite face. This rectangular prism cannot have more area than the parallelepiped, so $ 2A+h(2a+2b)\le 216\Rightarrow A+h(a+b)\le 108$ Therefore $ 108\ge A+h(a+b)\ge A+h(2\sqrt{A})=A+2\frac{216}{\sqrt{A}}\ge 108$. From this, you must have equality in all inequalities, so it's a cube.

I haven't worked out the details but shouldn't this be doable with AM-GM or something?

I don't think that works until you can establish that it's a rectangular prism. Of course, once you establish that, then it's just: $ abc=216 \implies \frac{a^{\frac{1}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}}}{6}=1$ $ 2(ab+bc+ca)=abc \iff \frac{a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{1}{3}}+a^{\frac{4}{3}}b^{\frac{1}{3}}c^{\frac{4}{3}}+a^{\frac{1}{3}}b^{\frac{4}{3}}c^{\frac{4}{3}}}{3}=abc$ $ \implies a=b=c$

Let $a, b, c$ to be the sides of the parallelepiped. We have that $abc = 216 (I)$ and that $2(ab + bc + ca) = 216 (II)$. From $(I)$ and $(II)$ we get $\sum\dfrac{1}{a} = \dfrac{1}{2}$. Squaring both sides, we get $2\sum \dfrac{1}{ab} = \dfrac{1}{4} - \sum \dfrac{1}{a^2}$ $(*)$. By Cauchy, $2\sum \dfrac{1}{ab} \ge \dfrac{9}{54} \Leftrightarrow \dfrac{1}{4} - \sum \dfrac{1}{a^2} \ge \dfrac{9}{54}$ $\Leftrightarrow \dfrac{9}{108} \ge \sum \dfrac{1}{a^2} \ge \sum \dfrac{1}{ab} \ge \dfrac{9}{108}$. So we get equality in AM-GM, what implies $a=b=c$.

I think this is the easiest way to solve the system : $$abc=216$$$$ab+bc+ca=108$$ Apply Newton's inequality and C.B.S. one time to get: $$108^2=(ab+bc+ca)^2 \ge 3abc(a+b+c) \ge 3abc \sqrt{3(ab+bc+ca)} =3 \cdot 216 \cdot 3 \cdot 6 = 54 \cdot 216 = 108^2$$ Hence we have equality everywhere and so $a=b=c$, $$\odot$$