Yes. Let $\mathbb{R}^{alg}$ be the field of real algebraic numbers. This is a real closed field which means that, for any statement of first order logic, using the symbols $0$, $1$, $+$, $\times$, $=$, $<$, that statement is true in $\mathbb{R}^{alg}$ if and only if it is true in $\mathbb{R}$.

The statement "The volume of $\mathrm{Hull}(x_1, x_2, \ldots, x_n)$ is $V$" is a first order statement in the coordinates of the $x_i$ and $V$. **Proof sketch** There are finitely many simplicial complexes on the abstract vertex set $1$, $2$, ..., $n$. For each of these simplicial complexes, the statement that this simplicial complex is a triangulation of $\mathrm{Hull}(x_1, x_2, \ldots, x_n)$ is finitely many polynomial inequalities and, if the simplicial complex is such a triangulation, then the volume of $\mathrm{Hull}(x_1, x_2, \ldots, x_n)$ is a polynomial in the coordinates of the $x$'s. So we can encode $\mathrm{Vol}(\mathrm{Hull}(x_1, \ldots, x_n)) = V$ as

$\Delta_1$ encodes a triangulation of the convex hull and $V=\cdots$
OR $\Delta_2$ encodes a triangulation of the convex hull and
$V=\cdots$ OR ...

So I can talk about $\mathrm{Vol}(\mathrm{Hull}(x_1, \ldots, x_n))$ in the first order theory of ordered fields. Consider the statement:

There exists an $A$ such that there are $n$ points on the unit sphere
with $A = \mathrm{Vol}(\mathrm{Hull}(x_1, \ldots, x_n))$ and such
that, for any points $y_1$, \dots, $y_n$ on the unit sphere, we have
$A \geq \mathrm{Vol}(\mathrm{Hull}(x_1, \ldots, x_n))$.

This is a first order statement by the above discussion. It is true when all the variables range over $\mathbb{R}$, because $(S^2)^n$ is compact. So it is also true when all variables range over $\mathbb{R}^{alg}$. Let $A^{alg}$ be the maximum volume of a sphere with real algebraic coordinates; we have just shown that this number exists. Let $A$ be the maximum volume of a sphere with real coordinates.

It remains to check that $A = A^{alg}$. Let $f(t)$ be the minimal polynomial of $A^{alg}$ over $\mathbb{Q}$. Let the real roots of $f$ be $x_1$, $x_2$, ...., $x_n$ with $A^{alg} = x_r$. Consider the statement

The number $A$ in the previous paragraph obeys $f(A) = 0$. There are
$n-1$ other numbers $x_1$, $x_2$, ..., $x_{r-1}$, $x_{r+1}$, ...,
$x_n$ with $f(x_1)=f(x_2) = \cdots = f(x_n)=0$ and $x_1 < x_2 < \cdots
> < x_{r-1} < A < x_{r+1} < \cdots < x_n$.

This is a first order statement which has been constructed to be true in $\mathbb{R}^{alg}$. So it is also true in $\mathbb{R}$, and we see that $A$ is the $r$-th root of $f$ in $\mathbb{R}$. Thus, $A = A^{alg}$, as claimed.