What kind of solid has this volume? In a certain application, I came across the $n$-fold integral:
$$
J_n = \int_{\theta_1 \in [0,2\pi]} \int_{\theta_2 \in [\theta_1, 2\pi]} \ldots \ldots \int_{\theta_i \in [\theta_{i-1},2\pi]} \ldots  \int_{\theta_n \in [\theta_{n-1},2\pi]} d\theta_{n}\ldots d\theta_{2}d\theta_{1} =
\frac{ (2\pi)^{n} }{ n!}
$$

What is the $n$-dimensional solid whose volume is $J_n$? 

For $n=3$, $J_n$ is the volume of a torus with both radii set to unity. Does this hold for larger $n$?

This looks to me like a very misleading setup. Note that the number $2\pi$ has nothing really to do with the problem. If you replace that with, say, $L$, then the evaluation is the same, your integral is $L^n/n!\,$. So my recommendation to you is to clear your mind of figures of revolution and look at the problem ab initio. When you do this, you see that the volume calculated belongs to an $n$-simplex: in case $n=2$, the right triangle with vertices $(0,0)$, $(0,L)$, and $(L,L)\,$; in case $n=3$, the “right tetrahedron” with vertices $(0,0,0)$, $(0,0,L)$, $(0,L,L)$, and $(L,L,L)\,$.
