While I do not object to the practical impossibility (or impracticality) of a Monte Carlo resolution of this problem with a precision of 6 decimal places pointed out by whuber, I would think a resolution with six digits of accuracy can be achieved.
First, following Glen_b, the particles are exchangeable in a stationary regime, hence it is sufficient (as in sufficiency) to monitor the occupancy of the different cells, as this constitutes a Markov process as well. The distribution of the occupancies at the next time step $t+1$ is completed determined by the occupancies at the current time $t$. Writing the transition matrix $K$ is definitely impractical but simulating the transition is straightforward.
Second, as noted by shabbychef, one can follow the occupancy process on the 450 odd (or even) squares, which remains on the odd squares when only considering even times, i.e. the squared Markov matrix $K^2$.
Third, the original problem only considers the frequency of zero occupancies, $\hat{p}_0$, after $50$ Markov transitions. Given that the starting point has a very high value for the stationary probability distribution of the Markov chain $(\mathbf{X}^{(t)})$, and given that focus on a single average across all cells,$$\hat{p}_0=\frac{1}{450}\sum_{i=1}^{450}\mathbb{I}_0(X_i^ {(50)})$$we can consider that the realisation of the chain $(\mathbf{X}^{(t)})$ at time $t=50$ is a realisation from the stationary probability distribution. This brings a major reduction to the computing cost, as we can simulate directly from this stationary distribution $\mathbf{\pi}$, which is a multinomial distribution with probabilities proportional to 2, 3, and 4 on the even corner, other cells on the edge, and inner cells, respectively.
Obviously, the stationary distribution provides directly the expected number of empty cells as
$$\sum_{i=1}^{450} (1-\pi_i)^{450}$$
equal to $166.1069$,
pot=rep(c(rep(c(0,1),15),rep(c(1,0),15)),15)*c(2,
rep(3,28),2,rep(c(3,rep(4,28),3),28),2,rep(3,28),2)
pot=pot/sum(pot)
sum((1-pot)^450)-450
[1] 166.1069
which is quite close to a Monte Carlo approximation of $166.11$ [based on 10⁸ simulations, which took 14 hours on my machine]. But not close enough for 6 decimals.
As commented by whuber, the estimates need to be multiplied by 2 to correctly answer the question, hence a final value of 332.2137,