It can be written using a hypergeometric function
$${\mbox{$_{n+1}$F$_n$}\left(1,3,\ldots,3;\,\frac32,\ldots,\frac32;\,{\frac{1}{4^n}}\right)}$$
I don't know if further simplification is possible.
EDIT: As requested, here is some elaboration. By definition, this generalized hypergeometric function $f$ is
$$ \sum_{k=0}^\infty \frac{(1)_k \left((3)_k\right)^n}{k!\; \left((3/2)_k\right)^n} (1/4)^{nk}$$ using the pochhammer symbols $$(z)_k = \Gamma(z+k)/\Gamma(z)$$ Thus $(1)_k = k!$, $(3)_k = (k+2)!/2$, $$(3/2)_k =\frac{\Gamma(3/2+k)}{\Gamma(3/2)} = \prod_{j=1}^{k} \left(\frac{2j+1}{2}\right) = \frac{(2k+1)!}{k!\; 4^k}$$ and
$$ f = \sum_{k=0}^\infty \left(\frac{(k+2)! k! }{2\;(2k+1)!}\right)^n = \sum_{k=0}^\infty \frac{1}{(C_{k+1})^n}$$