$\newcommand\si{\sigma}$ $\newcommand\Si{\Sigma}$ $\newcommand\R{\mathbb R}$ Let $\Si:=\{\pm 1\}^n$. The map $$\R^{n\times n}\ni w\mapsto f(w):=(w_\si)_{\si\in\Si}\in\R^\Si, $$ where $w_\si:=\si^T w\si$, is linear. Therefore and because $W$ is zero-mean Gaussian, we see that $$(W_\si)_{\si\in\Si}:=f(W):=f\circ W$$ is zero-mean Gaussian, with $W_\si:=\si^T W\si=\sum_{i,j}\si_i\si_j W_{ij}$; everywhere here, the summation indices run over the set $\{1,\dots,n\}$. Also, for all $\rho$ and $\si$ in $\Si$ $$EW_\rho W_\si=\sum_{i,j,k,l}\rho_i\rho_j\si_k\si_l EW_{ij}W_{kl} \\ =\sum_{i,j,k,l}\rho_i\rho_j\si_k\si_l 1_{\{i,j\}=\{k,l\}} \\ =\sum_{i,j,k,l}\rho_i\rho_j\si_k\si_l 1_{i=j=k=l}\\ +\sum_{i,j,k,l}\rho_i\rho_j\si_k\si_l 1_{i=k\ne j=l} \\ +\sum_{i,j,k,l}\rho_i\rho_j\si_k\si_l 1_{i=l\ne j=k} \\ =2\sum_{i,j,k,l}\rho_i\rho_j\si_k\si_l 1_{i=k,j=l} \\ -\sum_{i,j,k,l}\rho_i\rho_j\si_k\si_l 1_{i=j=k=l} \\ =2\sum_{i,j}\rho_i\rho_j\si_i\si_j -\sum_i\rho_i\rho_i\si_i\si_i \\ =2(\rho\cdot\si)^2-n, $$ where $\rho\cdot\si:=\sum_i\rho_i\si_i$; in particular, $EW_\si^2=EW_\rho^2=2n^2-n$. So, $$E(W_\si-W_\rho)^2=EW_\si^2+EW_\rho^2-2EW_\rho W_\si =4n^2-4(\rho\cdot\si)^2\le4n^2. $$ So, the bound you are getting is actually $2\sqrt{\log2\,}\, n^{3/2}$, $\sqrt2$ times as muchlarge as you suggested.