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-...
That is the entropy rate. If I'm understanding your original question correctly, you were asking about the standard equilibrium-defining thermodynamic entropy?