Anthony's comment has the correct answer: $Z=(z_1,\ldots ,z_N)$ is uniformly distributed on $\|Z\|^2=N$, $\langle e, Z\rangle=0$, $e=(1,1,\ldots, 1)$ (that is, the joint distribution is the $(N-2)$-dimensional Hausdorff measure restricted to this set and normalized).
As $Z$ is obviously supported by this set, this follows from the fact that the distribution is invariant under rotations $R\in O(n)$ with $Re=e$ and under $Z\mapsto -Z$. To see this, recall that $Y=RX$ has the same joint distribution as $X$. See here.
Moreover, $$ N\overline{Y}=\langle e, Y\rangle =\langle e, RX\rangle = \langle R^te, X\rangle = \langle e, X\rangle = N\overline{X} $$ and similarly $NS_Y^2=\|Y-\overline{Y}e\|^2=\|R(X-\overline{X}e)\|^2=NS_X^2$. Thus $$ RZ = \frac{1}{S_X}R(X-\overline{X}e) = \frac{1}{S_Y}(Y-\overline{Y}e) , $$ and indeed $Z$ and $RZ$ have the same distribution.