I got an alternative approach to prove the upper bound $J_n \leq C^n$, based on Fedja's idea of covering simplexes by cubes, and on David's description of $J_n$ using the functions $f$.

Using a change of variable, we convert all the $n!$ integrands in $J_n$ to be the same, namely $1/\sqrt{x_1x_2\cdots x_n}$. Now the integrands are $n!$ (possibly different) polytopes {$P_i:\;i=1,\cdots,n!$}. We now cover each $P_i$ by small cubes of side length $1/n$. The number of cubes required is about $Vol(P_i) n^n$, where $Vol(P_i)$ is the volume of $P_i$. The small cube with corner at the origin contributes most to the integral.

After a change of variable $x_j=y_j/n$, it remains to prove $$\sum_{i=1}^{n!}Vol(P_i)\leq C^n$$

This maybe as hard as the original question, but
{$P_i:\;i=1,\cdots,n!$} can be described by the collection of matrices {$M_f: f$}, by a 1-1 correspondence. Recall that $f:\{ 2,3,\ldots, n+1 \} \to \{1,2,\ldots, n \}$ obeys $f(i) < i$.

Given such a function $f$, we create an upper triangular $n\times n$  matrices matrix $M_f$ as follows: All entries of $M_f$ is either 0 or 1, and on the $j-$th column, an entry $M(i,j)$ is 1 if and only if $f(j-1) \leq i\leq j$.


For example, when $n=3$ and $(f(2),f(3),f(4))=(1,2,1)$, then the $M_f=$
$\begin{matrix}
1 0 1, \\
0 1 1,\\
0 0 1
\end{matrix}$

(These are the 3 rows of $M$ in the correct order, the output does not show the matrix in this environment and I don't know why)

These row vectors of $M_f$, together with the origin, gives rise to the polytope in $\{P_i:\;i=1,\cdots,6\}$.