Suppose we have a random tree where the probability that a node has $n$ successors is given by $\delta(n)$. What is the distribution of the number of nodes at the $s$-th level deep in the tree, $N_s(n)$?
Clearly $N_1(n) = \delta(n)$.
To calculate $N_2(n)$, we could do something like the binomial construction and have $N_2(n) = \delta(n_1) \delta(n_2) F(n, n_1, n_2)$, where $F(n, n_1, n_2)$ something like the choose function, in this case the number of ways that $n = n_1 n_2$.
But how to proceed from here?
An alternative approach, as indicated in my previous comment, is to interpret the random-tree as a branching process:
Let $X_n$ be the number of nodes in generation (depth) $n$. It is standard to show that the generating function of$X_n$ fulfills the following recursion
$G_{n+1}(z) = \mathbb{E}[z^{X_{n+1}}] = G_n(G(z))$
where $G(z) = \mathbb{E}[z^{X}] = \sum_{k\geq 0} \delta(k) z^k$.
$\delta(k)$ is the local offspring probability mass function.
Then once this recursion is solved you can use $\mathbb{P}[X_n = m] = G^{(m)}_n(0)/m!$, here $G^{(m)}$ is the $m$'th derivative w.r.t. $z$.
See for example the chapter about Branching Processes in Grimmett/Welsh "Probability: An Introduction"