We first note that \(2n^2 + 23n + 11\) factors as \((2n + 1)(n+ 11)\) . (We can find these factors using the rational root theorem.) Thus we have \( \lvert 2n^2 + 23n + 11 \rvert = \lvert 2n + 1 \rvert \cdot \lvert n + 11 \rvert . \)Now, each of the factors on the right hand side of this equation is an integer. It follows that the left hand side is a prime number if and only if one of the right hand factors is 1 and the other one is a prime number. Thus we must either have \(2n + 1 = \pm 1\) , or \(n +11 = \pm 1\). We consider these cases separately.

If 2n+1, then n=0, and n+11=11, which is prime. Thus this value of n works.

If 2n+1=-1 , then n=-1, and n+11=10 , which is not prime. Therefore we have no solution in this case.

If n+11=1 , then n=-10 , so 2n+1=-19 . Since 19 is prime, we obtain a valid solution in this case.

Finally, if n+11=-1, then n=-12 , and 2n+1=-23. Since 23 is prime, this value of n works.

Thus there are exactly \(\boxed{3}\) values of \(n\) that work: 0, \(-10\), and \(-12\) ; and these give the prime numbers 11, 19, and 23.