As Pat Devlin commented, you can compute the expected value. A similar calculation gives you the higher moments.
$$E(Y^2) = \sum_{i,j} I_{X_i = 0} I_{X_j = 0} \\\ = \sum_i I_{X_i = 0} + 2 \sum_{i\lt j} I_{X_i = 0} I_{X_j = 0} \\\ = \sum_i (1-p_i)^n + 2\sum_{i \lt j} (1-p_i-p_j)^n$$
$$E(Y^m) = \sum_{S \subset \lbrace 1, 2, ... n \rbrace} \\# surj(m,|S|) (1-P(S))^n$$
where $\\#surj(a,b)$ is the number of surjections from a set of size $a$ to a set of size $b$, $\sum_i (-1)^i {b \choose i }(b-i)^a$.