Use inclusion-exclusion. There are $4^6$ sequences altogether. For each of the $4$ letters there are $3^6$ sequences that don’t contain that letter, so subtract $4\cdot3^6$. Unfortunately, some sequences miss two letters, and we’ve subtracted each of these twice. There are $\binom42=6$ pairs of letters, and for each pair there are $2^6$ sequences that miss both letters, so we have to add back in $6\cdot2^6$. Finally, there are $4$ sequences that contain only one of the letters; we subtracted each of these three times and added them back in three times, so we need to subtract them one more time. The final result is
$$4^6-4\cdot3^6+6\cdot2^6-4=1560\;.$$
If you know the inclusion-exclusion formula, you can write this directly as
$$\sum_{k=0}^4(-1)^k\binom4k(4-k)^6\;.$$