If I am understanding the problem correctly, it appears that we can express everything in terms of the number $X_n$ of 1s in the array at step $n$. This gives a much simpler Markov chain to analyze.
Let $N = 1000$, and suppose at time $n$ we have $k$ 1s in the array. Suppose first that $A[j_1] = 0$ (happens with probability $(N-k)/N$). We set it to 1, gaining one 1 (so now there are $k+1$ 1s). If $A[j_2]$ and $A[j_3]$ are both 1 (prob $((k+1)/N)^2$) or both 0 (prob $((N-(k+1))/N)^2$), no more 1s are gained. If exactly one of $A[j_2], A[j_3]$ is 1 (prob $2(k+1)(N-(k+1))/N^2$), an additional 1 is gained.
Similarly, if $A[j_1] = 1$ (prob $k/N$) then we gain an additional 1 with probability $k(N-k)/N^2$, or no additional 1s with probability $(k^2 + (N-k)^2)/N^2$.
So putting this all together, we have the transition probabilities for $X_n$ are: $\begin{align*} p(k, k) &= \frac{k(k^2 + (N-k)^2)}{N^3} \\ p(k, k+1) &= \frac{k^2(N-k) + 2 (N-k)(k+1)(N-(k+1))}{N^3} \\ p(k, k+2) &= \frac{(N-k)((k+1)^2 + ((N-(k+1))^2)}{N^3} \end{align*} $
Perhaps someone can double-check that these sum to 1, as I haven't the time right now. But this should at least reduce the problem. There might be a slick martingale solution; I'll edit if I think of something.