Firstly, let us note that $5^3 - 4^3 = 61$, $2^3 - 1^3 = 7$ and $3^3 - 2^3 = 19$ are nice primes, so 1, 7 and 9 belong to desired digits. We show that they are all desired digits.
Let $p = m^3 - n^3$ be a nice prime, where $m > n$ are positive integers. Second factor in rewriting
$$p = m^3 - n^3 = (m-n)(m^2 + mn + n^2),$$
is greater than 1, thus the first one is 1 and therefore $m = n + 1$. After substitution we obtain
$$p = 3n^2 + 3n + 1. \qquad (1)$$
An estimate $3n^2 + 3n + 1 > 6$ gives that the prime $p$ is odd and greater than 5. This excludes 0, 2, 4, 5, 6 and 8 as the last digits and 3 stays the only remaining digit to exclude.
It is sufficient to find remainders of the numbers $3n^2 + 3n + 1$ after division by 5. For remainders 0, 1, 2, 3 and 4 of $n$ we obtain remainders 1, 2, 4, 2, 1 of (1) which ones really exclude the last digit 3.

*Answer.* The last digits of the nice primes are 1, 7 and 9.