Even with a distinguished root and insisting that all isomorphisms respect this
distinguished root, this will be a challenging enumeration.

For $k \leq 2$ the problem is straightforward: The count is the number of
partitions of $d_2$ into at most $d_1$ parts.  This has an upper bound of
$\binom{d_2+d_1 - 1}{d_1 - 1}$, call this quantity $q(d_2,d_1)$, and the literature doubtless has more
to say on the exact count.  Call this number $p(d_2,d_1)$, and let such a
partition be denoted by the vector $(n_1,n_2,\ldots, n_{d_1})$, where the
$n_i$ are in decreasing order and sum to $d_2$.  To handle $d_3$ will
involve a summation over all such partitions of products of terms like 
$p(k_i,n_i)$, where the $k_i$ sum up to $d_3$.  Except it won't be that
simple, as you need $k_i$ to be 0 when $n_i$ is 0, and you need to
identify certain counts when you have $n_i=n_{i+1}$, and so on.
Using $\prod q(d_{i+1},d_i)$ as an upper bound will likely be a weak
estimate, unless all the $d$'s are small, and even then I would compare
with a computer enumeration.

Of course $n(c,d_2,d_1)$ will often be much smaller than $p(d_2,d_1)$,
where this new count restricts the parts $n_i \lt c$, and there should be
some literature on $n()$ as well.  I still recommend computer enumeration
for cases of small diameter.