There are $4^{12}$ possibilities if there are no restrictions.
But there are exceptions. There are $3^{12}$ possibilities that are missing a given letter, and there are 4 letters, so we want to take away $4 \cdot 3^{12}$.
However, now it gets tricky. The four sets we subtracted have overlaps and we need to figure out those overlaps. We have to consider all the ones here we have double/triple counted and "took away" more than once.
If a string is missing two letters (i.e., both A and B), it will be counted twice. There are $2^{12}$ possibilities missing both A and B, $2^{12}$ possibilities missing both A and C, $2^{12}$ possibilities missing both A and D, $2^{12}$ possibilities missing both B and C, $2^{12}$ possibilities missing both B and D, and $2^{12}$ possibilities missing both C and D. All in all, there are $6 \cdot 2^{12}$ possibilities for missing two letters.
Lastly, we have to consider the final case of strings missing three letters. There are only 4 of them (all A's, all B's, all C's, and all D's). All A's was counted three times -- once under "missing C+D", once under "missing B+D", and once under "missing B+C", so we have to take away 2 of these copies. We had mistakenly added "all A's" back three times. Same for the other 3 letters. So minus $4 \cdot2=8$ here.
Thus, the final answer is
$4^{12} - 4 \cdot 3^{12} + 6 \cdot 3^{12}-4\cdot2 = 14676020$
I probably made a mental error in there somewhere, please correct me if you see one.