It's very likely true that there are infinitely many prime values of $p(n)$, but as usual, this is very unlikely to be provable with our current understanding. The usual heuristic for the number of prime values of a sequence $a_n$ with $n < x$ is $$\sum_{n<x} \frac{1}{\log(a_n)},$$ which in this case diverges because the asymptotic formula $$\log(p(n)) \sim \pi \sqrt{2n/3}$$ (reference here) implies $$\sum_{n<x} \frac{1}{\log(p(n))} \sim \frac{\sqrt{6x}}{\pi},$$ which diverges as $n \to \infty$. This heuristic needs some mild corrections due to known congruence properties of $p(n)$ due to Ramanaujan, Atkin, Ono, etc. However, numerical evidence supports the claim that it is at least on the right order of magnitude: according to the OEIS, there are $4967$ values of $n < 10^8$ such that $p(n)$ is prime, whereas the naive heuristic would predict $$\sum_{n<10^8} \frac{1}{\pi \sqrt{2n/3}} \approx 7796,$$ and correcting for the three congruences known to Ramanaujan gives $$\sum_{n<10^8} \frac 45 \cdot \frac 67 \cdot \frac{10}{11} \cdot \frac{1}{\pi \sqrt{2n/3}} \approx 4860.$$
As a side note: Peter mentions in a comment that it is surprising that the largest known prime value of $p(n)$ has only $40000$ digits, and that perhaps this indicates there are no larger ones up to a reasonable limit. The actual explanation for this is less exciting: since we cannot ordinarily factor $p(n) \pm 1$, we have no faster method than elliptic curve primality proving to prove the primality of $p(n)$, even when it appears to be prime. (ECPP is currently only practical up to several tens of thousands of digits; that $40000$-digit prime previously held the record. See discussion here and here.) Indeed, there are several larger values of $p(n)$ (see here) that are known to be probable primes.
