Flajolet and Sedgewick, Analytic Combinatorics (link goes to free, legal downloadable PDF of book), section VIII.6 treats the asymptotics of various types of partitions. They get that the number of partitions of $n$ into prime parts, which they denote $P_n^{(\Pi)}$ (and which is A000607) satisfies $\log P_n^{(\Pi)} \sim (2\pi/\sqrt{3}) \sqrt{n/\log n}$ (Here $f(n) \sim g(n)$ has the usual meaning $\lim_{n \to \infty} f(n)/g(n) = 1$.)

I believe (but have not explicitly checked) that if you disallow any finite number of primes this asymptotic formula still holds; in particular it should hold in the case you're asking about.

**Edited to add**: The logarithmic growth rate comes from a saddle-point estimate which can in turn be derived from the rate of growth of $\prod_{n \ge 1} 1/(1-z^{p_n})$ (where $p_n$ is the $n$th prime) as $z \to 1^-$ along the real axis. As stated after the equation (73) in F&S, p. 580,
$$ \sum_{n \ge 1} e^{-tp_n} \sim {t \over \log t} $$
and it's from this that the asymptotic result is derived. (Note that $z \to 1^-$ corresponds to $t \to 0^+$.) But if we omit the single term $e^{-2t}$ from that sum it won't change the asymptotic behavior at 0.

This isn't a full proof, though. The result is also a little bit counterintuitive, since one expects that a random partition of $n$ into primes, for large $n$, will contain at least one 2, and therefore leaving the 2s out should reduce the count considerably. But on the logarithmic scale this *appears* to be negligible.