Solution:

For even $n,$ the factors are five consecutive odd numbers. Among any five consecutive odd numbers, at least one is divisible by $3$ and exactly one by $5,$ so the product is always divisible by $15.$

No larger fixed divisor works: $n=2$ gives $3 \cdot 5 \cdot 7 \cdot 9 \cdot 11,$ whose greatest common divisor with other cases such as $n=10$ is exactly $15.$

Thus, the correct answer is D.