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 $\{P_i:\;i=1,\cdots,6\}$.

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 $n!$ polytopes {$P_i:\;i=1,\cdots,n!$} which has a one to one correspondence to the collection of matrices {$M_f: f$} where $f:\{ 2,3,\ldots, n+1 \} \to \{1,2,\ldots, n \}$ obeys $f(i) < i$.

We now describe the correspondence. 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 $\{P_i:\;i=1,\cdots,6\}$.

The job now is to show that these polytopes patch up around the origin at most $C^n$ times. An idea is to project the vertices (except the origin) of each polytope onto a sphere to form a spherical polygon, then check how may times these polygons patch up on the sphere. This maybe as hard as the original problem (high dimensional polytopes are hard to deal with) but the symmetry of the integrand .may simply the counting job. It'd be interesting to know if such an approach works.