Inscribed $n$-polytope with $2^n$ vertices of maximal volume The question is in the title:

Question: Which inscribed $n$-dimensional polytope (inscribed in the unit sphere) with $2^n$ vertices has the largest possible volume?

Is it the $n$-dimensional cube? If not, how much larger can its volume be?
 A: For $n=3$, the maximal volume polytope with 8 vertices is described in that paper.
Berman, J. D.; Hanes, K., Volumes of polyhedra inscribed in the unit sphere in (E^3), Math. Ann. 188, 78-84 (1970). ZBL0187.19604.The link is paywalled, but a summary with the optimal shape drawn is available in this answer.
It is combinatorially very different from a cube: it is simplicial, and each vertex has degree 4 or 5. The correct optimal polyhedron also has only $D_2$ symmetry.
Edit. An inscribed polytope which maximizes the volume given the number of vertices is always simplicial (see Lemma 1 in
Horváth, Ákos G.; Lángi, Zsolt, Maximum volume polytopes inscribed in the unit sphere, Monatsh. Math. 181, No. 2, 341-354 (2016). ZBL1354.52016.), so the answer is "no" as well in any dimension $>2$.
The gap is even quantitative. Let $\mu_n$ be the largest volume of an $n$-dimensional polytope with $2^n$ vertices inscribed in the ball of radius $\sqrt{n}$ (this normalization is better since it counterbalances the fact that the ball of unit radius has a tiny volume for large $n$). The cube gives the lower bound $\mu_n \geq 2^n$. By taking direct products, we have  $\mu_{m+n} \geq \mu_m\mu_n$, and therefore $\mu_{3n} \geq \mu_3^n \geq (2+\epsilon)^{3n}$, exponentially better that the cube. We cannot do much better since the upper bound $\mu_n \leq C^n$ for some constant $C$ holds trivially by comparing with the ball itself.
