The problem is elementary for $n=5$.

We may regard that case as a combination of two triangular pyramids sharing a base $\triangle ABC$. Then the volume is always bounded by one-third the area of the common base times the distance between the two remaining points $D,E$ (the latter is always greater than or equal to the sum of the pyramid altitudes). Then this bound is saturated and both factors maximized by placing $D$ and $E$ at opposite points ("poles") and distributing $A,B,C$ uniformly along the "equator" (a maximally symmetric triangular bipyramid).

Let's look more broadly here, as other broader answers have yet to appear.

We begin with this 2014 answer from Math Stack Exchange, which concentrates on four to eight points in three dimensions. Summarizing those results gives:

$n=4$ -- regular tetrahedron

$n=5$ -- triangular bipyramid, as above

$n=6$ -- regular octahedron

$n=7$ -- pentagonal bipyramid (this would seem to overcrowd the "equator" with five points, but plausible alternatives such as the capped octahedron would be worse)

$n=8$ -- not a cube, which isn't even close; rather a $D_2$-symmetry object with triangular faces and vertices of order $4$ and $5$.

The linked answer cites Ref. [1] for proof.

Horváth and Lángi[2] give a lemma that all solutions with four or more points in three dimensions are polyhedra with triangular faces; and the faces are also simplices in higher dimensions. The above claims conform with this, as do well-known (but AFAIK not rigorously proved) arrangements for larger $n$ such as the laterally tricapped trigonal prism for $n=9$ and the regular icosahedron for $n=12$. The triangular face requirement excludes the regular dodecahedron, which in any event has long since been beaten for $20$ points.

Reference [2] also considers higher dimensionalities. For $n=d+1$ points on the surface of a $d$-dinensional ball the optimum is the regular simplex. With $n=d+2$ the optimum is described as combining regular-simplex cross-sections in two complementary subsets of the dimensions. Going back to the triangular bipyramid with $n=5$ and $d=3$, the relevant cross-sections are an equilateral triangle (the base of the bipyramid) in two of the dimensions, and a line segment (the axis) in the remaining dimension. With higher dimensionality the two orthogonal cross-sections should have as nearly equal dimensionalities as possible. Thus for $(n,d)=(6,4)$ we would select an equilateral triangle in two dimensions and another equilateral triangle in the remaining two dimensions, not a three-dimensional tetrahedron and a one-dimensional line segment; so the $(6,4)$ optimum is not a higher-dimensional bipyramid. For $n=d+3$ with $d$ odd the optimum is guaranteed by using three orthogonal sim-plectic cross-sections, such as the line segments which are diagonals of the regular octahedron for $n=6, d=3$.

**References**

Joel D. Berman, Kit Hanes (1970), "Volumes of polyhedra inscribed in the unit sphere in E³",
*Mathematische Annalen*, **188**, 1, 78-84.

Horváth, Ákos G.; Lángi, Zsolt (2016), "Maximum volume polytopes inscribed in the unit sphere", *Monatsh. Math.* **181**, No. 2, 341-354. ZBL1354.52016.