For an $n$-by-$n$ unitary matrix $U$ and a permutation $\sigma\in S_n$, let $$w_\sigma=(-1)^\sigma\det(U^*)\prod_{i=1}^n U_{i,\sigma(i)}.$$ Is $\int_{U(n)}\mathrm{Re}(w_{\sigma_1})\mathrm{Re}(w_{\sigma_2})dU\ge 0$ for all $\sigma_1,\sigma_2\in S_n$?

The integral can be expanded to $$\frac{1}{4}\int_{U(n)} (w_{\sigma_1}+\bar w_{\sigma_1})(w_{\sigma_2}+\bar w_{\sigma_2})dU=\frac{1}{2}\int_{U(n)} w_{\sigma_1}w_{\sigma_2}+w_{\sigma_1}\bar w_{\sigma_2}dU.$$ Once the $\det(U^*)$ factors have been expanded as sums over permutations, this can be evaluated using Weingarten functions (see Collins 2003). The latter term is \begin{align}\int_{U(n)} w_{\sigma_1}\bar w_{\sigma_2}dU&=(-1)^{\sigma_1\sigma_2}\int_{U(n)}U_{1\sigma_1(1)}\cdots U_{n\sigma_1(n)}U^*_{\sigma_2(1)1}\cdots U^*_{\sigma_2(n)n}dU\\ &=(-1)^{\sigma_1\sigma_2}Wg(\sigma_2^{-1}\sigma_1,n).\end{align} By Novak 2010, the element $\sum_\sigma Wg(\sigma,n)\sigma$ of the group algebra $\mathbb{C}[S_n]$ can be written as a product of elements $(n+J_k)^{-1}$ of the form $\sum_\sigma (-1)^\sigma|a_\sigma|\sigma$, hence $(-1)^\sigma Wg(\sigma,n)\ge 0$. Computer experiments suggest the former term is nonnegative as well, but I haven't been able to prove it.

Note that $\int_{U(n)}\mathrm{Re}(w_\sigma)dU=1/n!$ by a symmetry argument (see here).

I can show that $$I_{\sigma_1,\sigma_2}=\frac{1}{n!(n+1)!}\sum_{\pi\in C}\sum_{\tau_1,\tau_2\in R}(-1)^{\sigma\pi} [\tau_1\pi\tau_2=\sigma],$$ where $C$ and $R$ are the Young subgroups $\mathrm{Sym}(1,\dots,n)\times\mathrm{Sym}(n+1,\dots,2n)$ and $\mathrm{Sym}(1,n+1)\times\cdots\times\mathrm{Sym}(n,2n)$ respectively and $\sigma=\sigma_1\oplus\sigma_2\in C$. Is there a way to link this with Carlo's formula or otherwise show it's nonnegative?

**Proof:** By expanding the $\det U^*$ factors and applying Weingarten functions,
\begin{align*}
I_{\sigma_1,\sigma_2}&=(-1)^{\sigma_1\sigma_2}\sum_{\pi_1,\pi_2}(-1)^{\pi_1\pi_2}\int_{U(n)}\prod_{i=1}^n U_{i\sigma_1(i)}U_{i\sigma_2(i)}U^*_{\pi_1(i)i}U^*_{\pi_2(i)i}dU\\
&=(-1)^{\sigma}\sum_{\pi\in C}(-1)^{\pi}\sum_{\tau_1,\tau_2\in R}Wg(\sigma^{-1}\tau_1\pi\tau_2,n)\\
&=(-1)^{\sigma}\sum_{\lambda\vdash 2n}\frac{\chi^\lambda(1)^2}{(2n)!^2s_{\lambda,n}(1)}\sum_{\pi\in C}(-1)^{\pi}\sum_{\tau_1,\tau_2\in R}\chi^\lambda(\sigma^{-1}\tau_1\pi\tau_2).
\end{align*}
For each $\lambda$, $\chi^\lambda$ is being summed over cosets of $R$ and the dual character $\chi^{\lambda'}=\mathrm{sgn}\cdot\chi^\lambda$ is being summed over cosets of $C$. The sum of a character $\chi$ over a coset of a subgroup $H$ is $0$ unless $\mathrm{res}_H\chi$ contains the trivial character with multiplicity at least $1$. By the Frobenius-Young correspondence, the only irreducible contained in both $\mathrm{ind}^{S_{2n}}_R 1$ and $\mathrm{ind}^{S_{2n}}_C \mathrm{sgn}$ is $\chi^\mu$ with multiplicity $1$, where $\mu=(2^n)$. Therefore,
$$c_\lambda:=\sum_{\pi\in C}(-1)^{\pi}\sum_{\tau_1,\tau_2\in R}\chi^\lambda(\sigma^{-1}\tau_1\pi\tau_2)=\begin{cases}0 & \text{if }\lambda\neq\mu\\ c_\mu&\text{if }\lambda=\mu.\end{cases}$$
Evaluating the above formula for the character of the regular representation yields
$$\chi^\mu(1)c_\mu=\sum_{\lambda\vdash 2n}\chi^\lambda(1)c_\lambda=c_{reg}=\sum_{\pi\in C}(-1)^{\pi}\sum_{\tau_1,\tau_2\in R}(2n)![\sigma^{-1}\tau_1\pi\tau_2=1].$$
It's easy to show with the hook-length formula that $\chi^\mu(1)=\frac{(2n)!}{n!(n+1)!}$ and even easier to show that there is only one semistandard Young tableau of shape $\mu$ with entries in $[n]$, so $s_{\mu,n}(1)=1$.