= Solution
Write the four cyclic factors as $A=M_{ij}$, $B=M_{jk}$, $C=M_{kl}$ and $D=M_{li}$. The <Gaussian moment pairing theorem> gives three terms, corresponding to $(AB)(CD)$, $(AC)(BD)$ and $(AD)(BC)$. We sum each term over $i,j,k,l$.
For $(AB)(CD)$, the entry <covariance> gives
$$
\begin{aligned}
\sum_{i,j,k,l}\mathbb E[AB]\mathbb E[CD]
&=\sum_{i,j,k,l}(\delta_{ik}+\delta_{ij}\delta_{jk})
(\delta_{ik}+\delta_{kl}\delta_{li})\\
&=n^3+2n^2+n.
\end{aligned}
$$
Indeed the four products leave respectively three, two, two and one free indices. A cyclic relabelling gives exactly the same sum for $(AD)(BC)$.
For the crossing pairing $(AC)(BD)$,
$$
\begin{aligned}
\sum_{i,j,k,l}\mathbb E[AC]\mathbb E[BD]
&=\sum_{i,j,k,l}(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk})
(\delta_{jl}\delta_{ki}+\delta_{ji}\delta_{kl})\\
&=n^2+3n.
\end{aligned}
$$
The first product allows $i=k$ and $j=l$ freely, whereas each other product forces all four indices to coincide. Combining the three pairings and the <trace> normalization,
$$
\boxed{\mathbb E[m_4(n)]
=\frac{2(n^3+2n^2+n)+(n^2+3n)}{n^3}
=2+\frac5n+\frac5{n^2}\ \longrightarrow\ 2.}
$$
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