Interpretation for the determinant of a stochastic matrix?

Is there a probabilistic interpretation for the determinant of a stochastic matrix (i.e. an $n \times n$ matrix whose columns sum to unity)?

Well one interpretation of a stochastic matrix is as a state transition matrix. So consider the state after the $n^\text{th}$ transition, $S=M^n$, and notice that $$\det(S) = \det(M^n) = \det(M)^n$$

Except in marginal cases when $\det(M) = 1$ or when the transitions don't converge, we expect that $\det(M^\infty) = \det(M)^\infty = 0$. So the closer the determinant is to $1$, the slower the transitions reach the steady state. The closer the determinant is to $0$, the faster the transitions reach steady state.

I can add a few things to DanielV's answer, but this is still very incomplete, and I hope the question will attract a better answer than this one.

We know that the magnitude of the determinant of a stochastic matrix must be between 0 and 1 inclusive. It's equal to 1 if and only if the matrix is a permutation matrix, with the determinant itself being equal to 1 for an even permutation, and -1 for an odd permutation.

On the other hand if the determinant is zero it means that some of the rows are linearly dependent, and thus that the matrix is not invertible. If we interpret it as a state transition matrix then this means that some information has been irretrievably lost about the initial state after just a single time step. For example, consider the matrix $$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1/2 & 1/2 \\ 0 & 1/2 & 1/2 \\ \end{pmatrix}. $$ This has determinant zero. If the corresponding Markov chain is started in state 2 then after a single time step this information has been completely lost: the state is now 2 or 3 (with probability $1/2$), and there's no way to guess, even probabilistically, whether its previous state was 2 or 3. On the other hand, if it's started in state 1 then this information is preserved indefinitely, so a determinant of zero only means that some information is immediately lost completely.

Note also that this system is not ergodic, so it will never converge to a unique equilibrium state. Thus, we can't strictly interpret a small determinant as implying that the system goes to equilibrium rapidly.

This suggests that the determinant can be interpreted as putting some kind of bound on how good the system is at preserving information about its initial state. But at the moment I'm not sure how to make this formal.

Finally, I want to note an interesting possible connection between Kolmogorov's criterion and determinants, although again I'm not quite sure what it means, if anything.

Kolmogorov's criterion is concerned with whether a Markov chain has the important property of being reversible. Define a loop as an ordered cyclic sequence of states $i_1, i_2, \dots i_n$, in which (we might as well say) no state is repeated. Kolmogorov's criterion says that for every possible loop, $$p_{i_1i_2}p_{i_2i_3}\dots p_{i_{n-1}i_n}p_{i_ni_1} = p_{i_1i_n}p_{i_ni_{n-1}}\dots p_{i_3i_2}p_{i_2i_1}. $$ That is, the probability of the system going around the loop in the forward direction is equal to it going around the same loop in the reverse direction.

The interesting thing is that determinants are also concerned with these loop probabilities. For example, the determinant of a $3\times 3$ stochastic matrix is given by $$ (p_{12}p_{23}p_{31}) + (p_{13}p_{32}p_{21}) - (p_{11})(p_{23}p_{32}) - (p_{22})(p_{13}p_{31}) - (p_{33})(p_{12}p_{21}) + (p_{11})(p_{22})(p_{33}), $$ where I've used parentheses to indicate that this is a sum of products of loop probabilities. It uses every possible loop at least once. (This is true in general.) The sign of the term depends on the number of loops involved, or equivalently, on whether the product of the loop lengths is odd or even. (The connection between loops and determinants was noted and developed by Richard Levins, but in the context of dynamical feedbacks in ecology, rather than Markov processes.)

However, while there does seem to be some kind of potential connection between determinants and Kolmogorov's criterion, I'm not sure exactly what it is. (It doesn't seem to be possible to state Kolmogorov's criterion in terms of determinants, for example.) It's suggestive that there might be quite a 'deep' interpretation of the determinant of a transition matrix, but we will have to wait for another answer to find out exactly what it is.