Clearly, all primes $p>7$ don't divide $n$. Clearly, $8 \mid p^2-1 \mid p^6-1$, but $16 \nmid p^6-1$ in general (just choose $p=11$). Now, for the odd part, clearly $9 \cdot 7 \mid p^6-1$ (since $\varphi(9)=\varphi(7)=6$), but we can find a prime $p \equiv 2 \mod 3^3 \cdot 7^2, p>7$ and then $p^6-1 \equiv 2^6-1 \equiv 63 \mod 3^3 \cdot 7^2$. Hence $n=2^3 \cdot 3^2 \cdot 7$.

Great solution @above! My solution was a bit different at the begining: Let a be the greatest prime divisor of n. Then, for p=a, n doesn't divide p^6-1 so a<11. The rest are pore or less the same.