3 primes

For $p^{2} + 2p - 8 = (p + 4)(p - 2)$ to be prime, one of $p + 4$ or $p - 2$ must equal $1$ . As $p$ must be prime the only option is that $p - 2 = 1 \Longrightarrow p = 3$ , which is prime, as is $p + 2 = 5$ . Thus the only such prime, and hence the desired sum, is $\boxed{3}$ .

Assume that $p$ is not equal to $3.$ The $p$ is either of the form $3k+1$ or $3k+2$ for some integer $k.$

Case 1: $p$ is of the form $3k+1$ . So $p+2=3k+3=3(k+1)$ , implying that $3$ divides $p+2$ which cannot happen.

Case 2: $p$ is of the form $3k+2$ . So $p^2+2p-8=(3k+2)^2+2(3k+2)-8$ which is also divisible by 3 which cannot happen.

This implies that the only possible value for $p$ is 3. This gives $p+2=5$ and $p^2+2p-8=7$ which are indeed primes. Thus the sum of all possible values of $p$ is $\boxed{3}$ .