It's all about number theory!

First note that for $n \le 0$ we have that $n^{3} - 8n^{2} + 20n - 13 \le 0,$ and hence cannot be prime. Thus we are only concerned with $n \in \mathbb{N}.$

Now upon observation we note that $n = 1$ is a zero of the given expression, so factoring out $(n - 1)$ gives us that

$n^{3} - 8n^{2} + 20n - 13 = (n - 1)(n^{2} - 7n + 13).$ So for this to be prime, we must have that either

• (i) $n - 1 = 1 \Longrightarrow n = 2,$ which gives us the prime $3,$ or

• (ii) $n^{2} - 7n + 13 = 1 \Longrightarrow n^{2} - 7n + 12 = (n - 3)(n - 4) = 0 \Longrightarrow n = 3,4,$ giving us primes $2,3.$

Thus the only values of $n$ that make the given expression prime are $2,3,4,$ which sum to $\boxed{9}.$

The polynomial can be written as a product of two integers: $P(n)=(n-1)(n^2-7n+13)$ . For this product to be a prime number, we need one of the factors to be either $+1$ or $-1$ , otherwise the product will be composite or 0.

Possibilities for $n-1=\pm 1$ are: $n=0$ or $n=2$ .

Possibilities for $n^2-7n+13=1$ are $n=\frac {7 \pm \sqrt{7^2 - 4×1×12}}{2}$ so that $n=3$ or $n=4$ .

For $n^2-7n+13=-1$ there are no solutions.

Checking each of these possibilities for primality:

• $P(0)=-13$ (no, primes are positive by definition)
• $P(2)= 3$ (yes)
• $P(3)= 2$ (yes)
• $P(4) = 3$ (yes)

The n's for which P(n) is a prime add up as $2+3+4=\boxed{9}$