This question is a followup to MSE2732980, where it is shown that $$ \mathcal{J}_2=\iint_{(0,1)^2}\frac{dx\,dy}{\sqrt{1-x^2}+\sqrt{1-y^2}}=\frac{\pi(4-\pi)}{4}.\tag{$n=2$}$$
It comes natural to wonder if there is a simple closed form, in terms of elementary functions and polylogarithms, for the generalized integral $$\begin{eqnarray*}\mathcal{J}_n&=&\int_{(0,1)^n}\frac{d\mu}{\sum_{k=1}^{n}\sqrt{1-x_k^2}}\\&=&\int_{0}^{+\infty}\left(\int_{0}^{+\infty}\sin(\theta)e^{-x\sin\theta}\,d\theta\right)^n\,dx\\&=&\left(\frac{\pi}{2}\right)^n\int_{0}^{+\infty}\left(L_{-1}(x)-I_1(x)\right)^n\,dx. \end{eqnarray*}$$
Remark 1: from the second integral representation for $\mathcal{J}_n$ we have that $\{\mathcal{J}_n\}_{n\geq 1}$ is a log-convex sequence, and it might be related to some probabilistic problem, like the probability that a random walk in $\mathbb{Z}^n$ returns to the origin.
Remark 2:
$$ \int_{0}^{1}\frac{du}{A+\sqrt{1-u^2}} = \frac{\pi}{2}-\frac{\text{arctanh}\sqrt{1-A^2}}{\sqrt{1-A^2}} A=\frac{\pi}{2}-\sum_{k\geq 0}\frac{A(1-A^2)^k}{2k+1}$$
hence the computation of $\mathcal{J}_2$ is equivalent to the computation of $\phantom{}_2 F_1\left(\tfrac{1}{2},\tfrac{1}{2};\tfrac{3}{2};1\right)$.
Remark 3: by the tangent half-angle substitution
$$\mathcal{J}_3=8\iiint_{(0,1)^3}\left(\sum_{k=1}^{3}\frac{1-u_k^2}{1+u_k^2}\right)^{-1}\prod_{k=1}^{3}\frac{1-u_k^2}{(1+u_k^2)^2}\,d\mu$$
and $\mathcal{J}_3$ might be related to Watson's triple integrals.
