Let $t$ be a natural number. For a unitary matrix $U$ let $U_{1,1}$ be the top left matrix element of $U$. I am trying to figure out the value of $\int |U_{1,1}|^{2t} dH(U)$ where $H$ is the Haar measure over the unitary group.

I know that this integral could be expressed as $$\sum_{\sigma,\tau\in S_t} \mbox{Wg}_d(\sigma \tau^{-1}) = t!\sum_{\sigma\in S_t} \mbox{Wg}_d(\sigma)$$ where $d$ is the implicit dimension the unitary group acts on $\mbox{Wg}_d$ is the $d$-dimensional Weingarten function, but I have no idea how to calculate this sum.

What I can say is that (applying Weingarten's theorem in reverse) $$\sum_{\sigma\in S_t} \mbox{Wg}_d(\sigma) = \int U_{1,2} U^\dagger_{2,1}\ldots U_{1,t+1} U^\dagger_{t+1,1} dH(U),$$ and for $d \gg t$ I think this should become well approximated by $$\int U_{1,2} U^\dagger_{2,1} dH(U) \ldots \int U_{1,t+1} U^\dagger_{t+1,1}dH(U)$$ (intuitively, having a large dimension makes the correlation between a small subset of matrix elements weaker). The value of $\int U_{1,2} U^\dagger_{2,1} dH(U)$ is exactly the expected absolute value of some element of a random complex unit vector, which again I am not sure how to calculate.