The question is in the title:



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> **Question:** Which inscribed $n$-dimensional polytope (inscribed in the unit sphere) with $2^n$ vertices has the largest possible volume?
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Is it the $n$-dimensional cube? If not, how much larger can its volume be?



Inscribed $n$-polytope with $2^n$ vertices of maximal volume
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)**](http://dx.doi.org/10.1007/BF01435416), Math. Ann. 188, 78-84 (1970). [ZBL0187.19604](https://zbmath.org/?q=an:0187.19604).The link is paywalled, but a summary with the optimal shape drawn is available in [this answer](https://math.stackexchange.com/a/988708/248217).


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**](http://dx.doi.org/10.1007/s00605-016-0949-2), Monatsh. Math. 181, No. 2, 341-354 (2016). [ZBL1354.52016](https://zbmath.org/?q=an:1354.52016).), 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.


