= Solution
Let $g_i=(X_{i1},\ldots,X_{in})$ be row $i$ of $X$, viewed as a real <vector>. These are <independent> standard <Gaussian> <random vectors>, and $(XX^T)_{ij}=g_i\cdot g_j$. The <trace> expression for the <empirical spectral measure> gives
$$
\widetilde m_k(n)
=n^{-k-1}\sum_{i_1,\ldots,i_k}
(g_{i_1}\cdot g_{i_2})\cdots(g_{i_k}\cdot g_{i_1}).
$$
We need the norm <moments>
$$
\mathbb E\lVert g_i\rVert^2=n,\qquad
\mathbb E\lVert g_i\rVert^4=n(n+2),\qquad
\mathbb E\lVert g_i\rVert^6=n(n+2)(n+4).
$$
For completeness, $S=\lVert g_i\rVert^2$ has the <chi-squared distribution> with $n$ degrees of freedom. By <independence> and the <Gaussian integral>, its <moment-generating function> is $\mathbb E e^{sS}=(1-2s)^{-n/2}$ for $s<1/2$. Differentiating at zero once, twice and three times proves these three identities.
For the first <moment>,
$$
\boxed{\mathbb E[\widetilde m_1(n)]
=n^{-2}\sum_i\mathbb E\lVert g_i\rVert^2=1.}
$$
For the second <moment>, if $i\ne j$, <conditional expectation> given $g_i$ gives $\mathbb E[(g_i\cdot g_j)^2\mid g_i]=\lVert g_i\rVert^2$, hence an unconditional value $n$. Separating equal and unequal row indices yields
$$
\boxed{\mathbb E[\widetilde m_2(n)]
=n^{-3}\{n\,n(n+2)+n(n-1)n\}=2+\frac1n.}
$$
For the third <moment>, the summand is $(g_i\cdot g_j)(g_j\cdot g_k)(g_k\cdot g_i)$. There are three disjoint cases.
* If $i,j,k$ are all different, average over $g_k$ first. Its <covariance matrix> is the identity, so the <conditional expectation> of the last two factors is $g_j\cdot g_i$. The summand's <expectation> is therefore $\mathbb E(g_i\cdot g_j)^2=n$. There are $n(n-1)(n-2)$ such triples.
* If exactly two indices agree, cyclic symmetry reduces the calculation to $i=j\ne k$. The product is $\lVert g_i\rVert^2(g_i\cdot g_k)^2$, whose <conditional expectation> given $g_i$ is $\lVert g_i\rVert^4$. Its <expectation> is $n(n+2)$, and there are $3n(n-1)$ such triples.
* If all three indices agree, the product is $\lVert g_i\rVert^6$, with <expectation> $n(n+2)(n+4)$. There are $n$ choices.
Consequently
$$
\begin{aligned}
\mathbb E[\widetilde m_3(n)]
&=n^{-4}\{n^2(n-1)(n-2)
+3n^2(n-1)(n+2)+n^2(n+2)(n+4)\}\\
&=5+\frac6n+\frac4{n^2}.
\end{aligned}
$$
The three limiting <expected values> are therefore
$$
\boxed{\lim_{n\to\infty}\mathbb E[\widetilde m_k(n)]
=\begin{cases}1,&k=1,\\2,&k=2,\\5,&k=3.\end{cases}}
$$
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