Here's what I've got:
For $x\in [0,\pi/n]$, we have $\sin(x)=x(1+O(1/n^2))$. Similarly, we have $\sin^n(x)=x^n(1+O(1/n))$. Let us write $f(x)=\frac{\sin(x)}{x}$. We have $|f(x)|\leq 1$, but for any $\epsilon>0$, $\exists \delta>0$ such that $f(x)\geq 1-\epsilon$ for $x\in [0,\delta]$
Then, for $x\in[0,\pi/n]$:
$$
\prod_{k=1}^n\frac{\sin(kx)}{\sin(x)}=n!(1+O(1/n))\prod_{k=1}^n f(kx)
$$
Now, we have
\begin{eqnarray*}
J(n)&=&\frac{2}{\pi}n!(1+O(1/n))\int_0^{\pi/n}\prod_{k=1}^n f(kx)\,dx\\
&\geq&n!(1+O(1/n))\int_0^{\delta/n}(1-\epsilon)^n\,dx\\
&\geq&\frac{2}{\pi}(n-1)!(\delta+O(1/n))(1-\epsilon)^n
\end{eqnarray*}
Similarly, we have
$$
J(n)\leq 2(n-1)!(1+O(1/n))
$$
We can combine these two estimates to get
$$
J(n)=(n-1)!(1+o(1))^n
$$
Of course, $(n-1)!=n!(1+o(1))^n$, so we can write
$$
J(n)=n!(1+o(1))^n
$$
It might be useful to use Stirling's formula to get
$$
J(n)=\left(\frac{n}{e}(1+o(1))\right)^n
$$
Using Mathematic, I computed $\left(\frac{J(500)}{499!}\right)^{1/500}=.99554\ldots$