Each die contributes an exponent vector in the primes $2,3,5$ (face $1\to(0,0,0),$ $2\to(1,0,0),$ $3\to(0,1,0),$ $4\to(2,0,0),$ $5\to(0,0,1),$ $6\to(1,1,0)$), and a product is determined by the sum of these vectors.

Fix the exponent of $5$ by requiring exactly $k$ dice to show $5,$ and put $m=n-k.$ For an exponent $b$ of $3,$ where $0\le b\le m,$ the possible exponents $a$ of $2$ are precisely $0,1,\ldots,2m-b.$ If $a\le b,$ use $a$ faces showing $6$ and $b-a$ showing $3.$ If $a\gt b,$ use $b$ faces showing $6$ and make the remaining $a-b$ factors of $2$ with faces $4$ and, if needed, one face $2.$ The bound $a\le 2m-b$ says this uses at most $m$ dice; fill unused dice with $1$'s. Thus the number of exponent pairs is $$\begin{gathered} \sum_{b=0}^{m}(2m-b+1)\\ {}=\frac{(m+1)(3m+2)}2. \end{gathered}$$

Summing over $k=0,1,\ldots,n$, equivalently over $m=0,1,\ldots,n,$ the number of distinct products is $$\begin{gathered} \sum_{m=0}^{n}\frac{(m+1)(3m+2)}2\\ {}=\frac{(n+1)^2(n+2)}2. \end{gathered}$$ For $n=11$ this is $\dfrac{12^2\cdot13}{2}=936,$ so $n=11.$

Thus, the correct answer is A.