$j(\tau)=J(e^{2i\pi \tau})$, since $J(q)\in q^{-1}+\Bbb{Z}[[q]]$ then $J^{-1}(1/s)=\sum_{k\ge 1} d_k s^k$ is in $\Bbb{Z}[[s]]$.
$1/J(q)$ is surjective from $|q|< 1$ to $\Bbb{C}$ and locally biholomorphic away from $1/J(q)=1/1728$ where it has a double point.
We get that $f(s)=J^{-1}(1/s)$ is locally analytic on $\Bbb{C}$ minus a branch point at $1/1728$ where it is analytic in $(s-1/1728 )^{1/2}$.
$f'(1/1728)=\sum_{k\ge 1} \frac{k\, d_k}{1728^{k-1}}$ doesn't converge, this follows from that $J^{-1}(1/s) = \sum_{n\ge 0} c_n (1-1728 s)^{n/2}$ (where $c_1\ne 0$) converges for all $\Bbb{C}$.
To get an upper bound on the $d_k$, let $m(y)=\sup_{\Im(\tau)>y} |1/j(\tau)|$. For $|s|<m(y)$ we'll have $J^{-1}(1/s)<e^{-2\pi y}$ so that $|d_k| \le m(y)^{-k-1} e^{-2\pi y}$ whenever $m(y)< 1/1728$. A numerical check shows that $m(2)< 10^{-5}$.
$J^{-1}(1/s) = \sum_{n\ge 0} c_n (1-1728 s)^{n/2}$, which is valid for all $s$, gives that $d_k = 1728^k \sum_{n\ge 0} c_n {n/2\choose k}$ but I don't see if it may give an asymptotic (perhaps a Tauberian theorem?)