OK, no, I'm wrong. The entropy of a Markov chain with stationary distribution v is [0]:

    -\sum v_i p_{i,j} \log(p_{i,j})

That is, the "entropy" of the transition matrix modified by the stationary distribution.

[0] https://math.stackexchange.com/questions/1040972/entropy-of-...

Here's a thought: once could derive the spectrum of the Markov transition matrix, and assign an entropy to each of the eigenvectors. The dominant eigenvector (highest entropy) would be the ergodic / stationary distribution, but it seems likely that each successive eigenvector would have a little less entropy. One could initialize the system in a "localized" state (very low entropy) and study the thermalization process as each of the low-entropy eigen-components decay away (exponentially, with rates proportional to the corresponding eigenvalue of the transition matrix) finally leaving the system in the high-entropy stationary distribution. The balance between the eigenvalues (exponential rates) and the entropies of respective eigenvectors would characterize the rate of entropy production (at different times) in the Markov chain!

Find the equilibrium probabilities from first eigenvector, use definition of entropy like you did. If it's not ergodic then you get subsystems with their own entropies. Second eigenvalue iirc basically characterizes dominant relaxation time.

This is all if you mean the equilibrium entropy of the underlying system, not the entropy rate.

This predates stochastic thermo quite a bit. Entropy of markov chain is at least 50 y/o, maybe more.