Number of real solutions of a random equation

I'll make an attempt. I think this reduces your problem to an integral, but I don't know if it can be found in closed form.

Let $d\mu(J)$ be the probability density on the matrices $J$; the expected value for the number of solutions to the equation will be given by $$ \int d\mu(J) \int d\vec{x}\ [\vec{x}=J\vec{\phi}], $$ where $\vec{\phi} = \arctan\vec{x}$. Change the order of integration to get $$ \int d\vec{x}\int d\mu(J)\ [\vec{x}=J\vec{\phi}], $$ then we can find the inner integral. The outer integral is taken over all real vectors $\vec{x}\in\mathbb{R}^n$.

The condition $\vec{x}=J\vec{\phi}$ consists of $n$ independent conditions, each over $n$ entries of $J$. Consider one row of that equation: $$ \int d\mu(\vec{J}_i) [x_i = \vec{J}_i \cdot\vec{\phi}]. $$ Because $J_{ij}$ are i.i.d. Gaussian, with mean 0, we can assume without loss of generality that $\vec{\phi}$ is pointing along, say, the axis of $J_{i1}$, which removes $J_{i2},\ldots,J_{in}$ from the integrand, and implies that $$ \int d\mu(\vec{J}_i) \left[x_i = J_{i1} \|\vec{\phi}\|\right] = (2\pi\sigma^2)^{-1/2}\exp\left (-\frac{x_i^2/\|\vec{\phi}\|^2}{2\sigma^2} \right). $$

Now, $$\int d\mu(J) [1_{\vec{x}=J\vec{\phi}}] = (2\pi\sigma^2)^{-n/2} \exp\left(-\frac{\|\vec{x}\|^2/\|\vec{\phi}\|^2}{2\sigma^2}\right), $$ and so the answer is $$ I = (2\pi\sigma^2)^{-n/2} \int d\vec{x}\ \exp\left(-\frac1{2\sigma^2}\frac{\|\vec{x}\|^2}{\|\arctan \vec{x}\|^2} \right). $$

Now, take the fact that $|\arctan x|\leq |x|$, and approximate the integral by $$ I \leq (2\pi\sigma^2)^{-n/2}e^{-1/(2\sigma^2)}\pi^n + \cdots, $$ where $\pi^n$ is the volume of the cube $[-\pi/2,\pi/2]^n$, and I'm assuming (though this seems a reasonable guess) that the contribution from outside the cube is asymptotically negligible. Then $$ \lim_{n\to\infty} \frac1n \log I < \frac12\log(\pi/2) - \log\sigma. $$ Unfortunately, I'm not at all certain that this particular bound is either tight or even correct.