We begin by factoring $ 336 $. $ 336 = 2^4 \cdot 3 \cdot 7 $. Because we are looking for phone numbers, we want four single digits that will multiply to equal $ 336 $. Notice that $ 7 $ cannot be multiplied by anything, because $ 7 \cdot 2 $ is $ 14, $ which is already two digits. So, one of our digits is necessarily $ 7 $. The factor $ 3 $ can be multiplied by at most $ 2, $ and the highest power of $ 2 $ that we can have is $ 2^3 = 8 $. Using these observations, it is fairly simple to come up with the following list of groups of digits whose product is $ 336: $ \begin{align*} &1, 6, 7, 8\\ &2, 4, 6, 7\\ &2, 3, 7, 8 \\ &3, 4, 4, 7\end{align*} For the first three groups, there are $ 4! = 24 $ possible rearrangements of the digits. For the last group, $ 4 $ is repeated twice, so we must divide by $ 2 $ to avoid overcounting, so there are $ \frac{4!}{2} = 12 $ possible rearrangements of the digits. Thus, there are $ 3 \cdot 24 + 12 = \boxed{84} $ possible phone numbers that can be constructed to have this property.