If you restrict your attention to random vectors, which have independent entries, then the only possible case is Gaussian. First, note that you can permute coordinates to deduce that they are in fact IID. Multiplying one coordinate (say $z_1$) by $i$ and $-i$ we can conclude that real and imaginary parts are identically distributed and symmetric. Now, apply transformation $(z_1, \dots z_n) \to (\frac{z_1+z_2}{\sqrt{2}}, \frac{z_1 - z_2}{\sqrt{2}}, \dots z_n)$, which gives as independence of $z_1+z_2$ and $z_1 - z_2$. Also their real and imaginary parts are independent, so we need a lemma:

Lemma. Let $X,Y$ be IID, symmetric random variables, such that $X-Y$ and $X+Y$ are independent. Then $X$ is Gaussian.

Proof. Let's compute characteristic function of X:
$$ \varphi_X(2t) = \varphi_{X+Y + X-Y} (t) = \varphi_X(t)^{4}$$
$\Psi(t) := \varphi_X(t)^\frac{1}{t^2}$ satisfies $\Psi(t) = \Psi(2t)$, so we have $A^{t^2} \leqslant \varphi_X(t) \leqslant 1$ for some $A\leqslant 1$ which means that around zero we have $\varphi_X(t) \geqslant 1 - \varepsilon t^2$ for some $\epsilon >0$ (or $\varphi_X(t)\equiv 1$), hence $\mathbb{E}X^2 < \infty$. Now let $X_n$ be a sequence of IID random variables such that $X_n \sim X$. Then it can be easily seen that $\varphi_{\frac{X_1 + X_2 + X_3 + X_4}{2}}(t) = \varphi_X(t)$ and, inductively, $\varphi_{\frac{X_1 + \dots + X_{4^{n}}}{2^{n}}} = \varphi_X(t)$. But $X$ has finite second moment, hence by CLT the LHS converges to the characteristic function of a Gaussian, and the result follows.

Actually, what I've just written above, is only sufficient to write $(z_1, \dots, z_n)$ in a form $\mathbf{X} + i \mathbf{Y}$, where $\mathbf{X}$ and $\mathbf{Y}$ are identically distributed standard (possibly rescaled) Gaussian vectors, but one can exploit unitary invariance, to prove that joint distribution is also Gaussian.