Here is a general comment. Let $G$ be a compact group and let $V$ be a (finite-dimensional, continuous, complex) representation of $G$. This data determines a locally finite directed graph, the *representation graph* $\Gamma$ of $G$ and $V$, as follows. The vertices of $\Gamma$ are the irreducible representations of $G$, and the number of edges from an irreducible representation $U_1$ to an irreducible representation $U_2$ is the multiplicity of $U_2$ in $U_1 \otimes V$.

Let $\chi_V$ denote the character of $G$.

**Claim:** The moment

$$\int_G \chi_V(g)^k \, dg$$

is the number of closed walks of length $k$ from the trivial representation to itself in $\Gamma$.

The idea of the proof is to look at the relationship between the decomposition of $V^{\otimes k}$ into irreducibles and the decomposition of $V^{\otimes (k+1)}$ into irreducibles. The application is that if $G$ is sufficiently nice then we can try to figure out what $\Gamma$ looks like and work from there.

Let's do $G = \text{O}(3)$ ($V$ the complexification of the standard representation) this way to demonstrate how the method works. First observe that we can immediately reduce to $G = \text{SO}(3)$, since the integral over $\text{O}(3)$ splits up into an integral over $\text{SO}(3)$ and over the other coset, and since the latter is $-1$ times the former (here we are using the fact that $3$ is odd) we get that

$$\int_{\text{O}(3)} \chi_V(g)^{2k+1} \, dg = 0$$

and

$$\int_{\text{O}(3)} \chi_V(g)^{2k} \, dg = \int_{\text{SO}(3)} \chi_V(g)^{2k} \, dg.$$

Recall that the irreducible representations of $\text{SO}(3)$ are precisely the odd-dimensional irreducible representations $V_1, V_3, V_5 \dots$ of its Lie algebra $\mathfrak{so}(3) \cong \mathfrak{su}(2)$, with $V = V_3$.

**Claim:** For $n \ge 3$, we have $V_n \otimes V_3 \cong V_{n-2} \oplus V_n \oplus V_{n+2}$.

This should be familiar from the representation theory of $\mathfrak{su}(2)$, and it gets us a very explicit description of the representation graph in this case: the vertices are labeled by the odd positive integers $1, 3, 5 \dots$ and for every vertex there are three edges, one from each vertex to its left neighbor, to itself, and to its right neighbor. (The special case $V_1$ is easy to handle since $V_1 \otimes V_3 \cong V_3$.)

As a corollary, closed walks from $1$ to itself of length $k$ are very close to being counted by the Motzkin numbers $M_k$ (A001006), which would be the correct answer if the representation graph had an additional edge from $1$ to itself. Some computations reveal that the actual sequence we get is A005043, which I've never seen before but which OEIS says are called the Riordan numbers. The sequence begins

$$1, 0, 1, 1, 3, 6, 15, 36, 91 \dots$$

so to get the original sequence of moments we wanted, we start with this sequence and replace every odd term with $0$, giving

$$1, 0, 1, 0, 3, 0, 15, 0, 91 \dots$$

This sequence appears in the OEIS with the zeroes removed as A099251.

For general compact connected Lie groups the representation graph should look a bit like the intersection of a Weyl chamber with the weight lattice, or something like that. I think the combinatorics gets much hairier as soon as the rank is higher than $1$ though.