Revised.
Suppose there are $N$ realizations of Gaussian process denoted as the vectors $z_{j} \in \mathbb{R}^{n}$ for $j = 1, \ldots, N$. Let $y$ be a random variable such that $y = \sum_{j=1}^{N}(Bz_{j})[i]$ where $B$ is a unitary matrix $B^* B = BB^* = I$. What is the variance of $y^{2}$?
On
With $Y=\sum_j B_{ij}x_j$, $$Y^2=\sum_{jk}B_{ij}B_{ik}x_jx_k,\,Y^4=\sum_{jkmn}B_{ij}B_{ik}B_{im}B_{in}x_jx_kx_mx_n.$$Now we just need to compute $\Bbb EY^4-(\Bbb EY^2)^2$. With $x_j\sim N(\mu_j,\,\sigma_j^2)$, $$\Bbb EY^2=\sum_{jk}B_{ij}B_{ik}(\mu_j\mu_k+\delta_{jk}^2\sigma_j^2)=(B\mu)_i^2+(B\sigma)_i^2.$$Similarly, $$\Bbb E Y^4=\sum_{jkmn}B_{ij}B_{ik}B_{im}B_{in}\Bbb E(x_jx_kx_mx_n)$$can be computed with this theorem. If my arithmetric's right, $$\Bbb EY^4=3(B\mu)_i^4+6(B\mu)+i^2(B\Sigma B^T)_{ii}+6(B\Sigma B^T)_{ii}^2,$$so$$\operatorname{Var}Y^2=-2(B\mu)_i^2(B\sigma)_i^2-(B\sigma)_i^4+6(B\mu)_i^2(B\Sigma B^T)_{ii}+6(B\Sigma B^T)_{ii}^2.$$
A unitary matrix transforms a Gaussian random vector to another Gaussian random vector with the same expectation and variance. In particular, each coordinate is distributed like a one-dimensional standard Gaussian random variable, which means that its variance equals $1$, if the original Gaussian vector was standard.