Volume of a shape whose boundary consists of portions of spheres symmetrically placed about the origin in $d\gg 1$ dimensions We are given a convex shape $S$ in the $d$-dimensional Euclidean space, whose boundary is formed by portions of $2d$ different spheres, one portion per sphere. The radius of each sphere is the same, $a+1$, and the spheres centers are $\pm a\textbf{e}_j$ for $1\le j\le d$, where $a\ge 0$. Hence, for $a>0$ each sphere passes through one and only one of the $2d$ points in the set $P:=\{\pm\textbf{e}_j | 1\le j\le d\}$.
Finally, for each point $\mathbf{p}\in P$, the portion of the sphere $\mathcal{S}_{\mathbf{p}}$ passing through $\mathbf{p}$ consists of all points of $\mathcal{S}_{\mathbf{p}}$ whose distance from $\mathbf{p}$ is smaller or equal to the distance between $\mathbf{p}$ and any point of the other $2d-1$ spheres.

Questions: How can we provide a tight lower bound of $a$ in terms of $d$ to have that $\frac{V(S)}{2^d}$ is a constant bounded away from $0$ when $d\to\infty$?
 A: If we set $a = \beta d^2$ for $\beta > 0$ then $$\frac{V(S)}{2^d} \to \exp(-1/(6\beta)).$$
The idea is to choose $x \in [-1,1]^d$ uniformly at random.  Then $\frac{V(S)}{2^d}$ is just the probability that $x \in S$.
Let $S_j^+$ (respectively $S_j^-$) denote the ball centered at $e_j a$ (respectively $-e_j a$) of radius $a+1$.  For a point $x \in [-1,1]^d$, note that $x \in S_j^+$ if and only if $$(x_j - a)^2 + \sum_{i \neq j} x_i^2 \leq (a + 1)^2$$
i.e. if and only if $$-2x_j a + \|x \|_2^2 \leq 2a + 1\,.$$
Arguing the same for $S_j^{-}$ we have that $x \in S_j^+ \cap S_j^-$ if and only if $$  \|x\|_2^2 \leq 2a(1 - |x_j|) + 1\,.$$
This implies that $x \in S = \bigcap_j (S_j^+ \cap S_j^-)$ if and only if $$\|x \|_2^2 \leq 2a (1 - \|x \|_\infty) + 1$$
i.e. if $$\frac{\|x\|_2^2}{d} \leq 2\frac{a}{d^2} \left (d (1 - \|x \|_{\infty}) \right) + \frac{1}{d}.$$
As $d \to \infty$ we have $\|x\|_2^2/d \to 1/3$, while $d(1 - \|x \|_\infty)$ converges to an exponential random variable $Y$ of mean $1$.  Thus we have $$\frac{V(S)}{2^d} \to P(1/3 \leq 2 \beta Y) = \exp(-1/(6\beta)).$$
