The idea of complementary counting is correct.
To count the cases that the product is NOT divisible by $4$, there can be at most one factor of $2$. So far so good.
However, it is not as simple as you stated. One of the dice can be 1,2,3,5, or 6---this is true, but you also need to choose that die, so you need to multiply by $6$. That is not all the problems here. When this die has number 1, 3, or 5, it is odd, same as the other dice, so there is a lot of overcounting here.
To count it in a clean way, consider the cases (1) All dice show odd numbers, and (2) One of the dice is 2 or 6, and all others are odd. Then you can subtract both cases from the total.