We need to count how many times the factor 3 appears in the prime factorization of 15!. This can be done by counting the multiples of 3, 9, 27, and so on, up to 15:

Multiples of 3: 15, 12, 9, 6, 3 (5 numbers) Multiples of 9: 15, 6 (2 numbers) Multiples of 27: None, since $27 > 15$

Thus, the total count of factors of 3 is 5 (from multiples of 3) + 2 (from multiples of 9) = 7.

Therefore, the maximum value of $m$, the power of 3 in the factorization of 15!, is 6, as the remaining factors contribute to $n$.

Hence, the correct answer is 6.