The answer is that $1$ is the only such number. It suffices to show that every prime $p$ divides $a_n$ for some positive integer $n$. Note that both $p=2$ and $p=3$ divide $a_2 = 2^2 + 3^2 + 6^2 - 1 = 48$.

Now we assume that $p \ge 5$. By Fermat's Little Theorem, we have $2^{p-1} \equiv 3^{p-1} \equiv 6^{p-1} \equiv 1 \pmod{p}$. Then
$$3 \cdot 2^{p-1} + 2 \cdot 3^{p-1} + 6^{p-1} \equiv 3 + 2 + 1 \equiv 6 \pmod{p},$$
or, $6(2^{p-2} + 3^{p-2} + 6^{p-2} - 1) \equiv 0 \pmod{p}$; that is, $6a_{p-2}$ is divisible by $p$. Because $p$ is relatively prime to $6$, $a_{p-2}$ is divisible by $p$, as desired.