This is a standard application of Markov chain convergence theory. You can abstract the room configuration as a graph, with 6 vertices having degree 1 (leaves), two side vertices degree 3 and central vertex degree 4. Let $\mu_i$ be the long-term probability of the mouse spending in room $i$, and let $p_{ij}$ be the transition probability from room $i$ to room $j$. Then one must have $$ \sum_i \mu_i p_{ij} = \mu_j$$

It's easy to see that $\mu_i \propto d_i$ where $d_i$ is the degree of room $i$.

This is a standard application of Markov chain convergence theory. You can abstract the room configuration as a graph, with 6 vertices having degree 2 (leaves), two side vertices degree 4 and central vertex degree 5: because of self-loops as pointed out in comment belwo. Let $\mu_i$ be the long-term probability of the mouse spending in room $i$, and let $p_{ij}$ be the transition probability from room $i$ to room $j$. Then one must have $$ \sum_i \mu_i p_{ij} = \mu_j$$

It's easy to see that $\mu_i \propto d_i$ where $d_i$ is the degree of room $i$ is one (projective) solution; to show uniqueness look up the proof of Perron-Frobenius Theorem.