If we let $(X_1, X_2, X_3)$ be the triple representing the frequencies of the sides from $n = 15$ rolls. Then we can model this as a Multinomial distribution where $p_1 = p_2 = p_3 = 1/3$. We seek to evaluate $E[X_1 X_2 X_3]$.
First, we note that via the law of total expectation, we can write:
$$ E[ X_1 X_2 X_3 ] = \sum_{x_1 = 0}^{n} E[x_1 X_2 X_3 | X_1 = x_1] P(X_1 = x_1) = \sum_{x_1 = 0}^{n} E[ X_2 X_3 | X_1 = x_1] x_1 P(X_1 = x_1) $$
We used properties of expectations to pull out the factor of $x_1$ is the second step. Two important facts we will use: first, the marginal distribution of $X_1$ is simply binomial with probability $p = 1/3$. So we have:
$$ P(X_1 = x_1) = \binom{n}{x_1} \left(\frac{1}{3}\right)^{x_1} \left(\frac{2}{3}\right)^{n - x_1} $$
Second, the conditional distribution of $(X_2, X_3)$ given $X_1 = x_1$ is also binomial with $n \to n - x_1$ and $p = 1/2$. We can see this by thinking about the fact that if you roll $n$ d3s and take away all the 1s, then the remaining dice are 2s and 3s with 50% chance each. Since the remaining must be 2s and 3s, once we fix the amount of 2s (or 3s) the remaining dice must be 3s (or 2s). So the above conditional expectation reduces is equivalent to "let $Y \sim \textrm{Binomial}(m, 1/2)$, what is $E[Y(m - Y)]$? We can evaluate this as:
$$ E[Y(m - Y)] = mE[Y] - EY^2 $$
Since $\mathrm{Var}(Y) = EY^2 - (EY)^2$, we can write:
$$ E[Y(m - Y)] = mEY - \mathrm{Var}\,Y - (EY)^2$$
Using facts about the binomial, we know $EY = m/2$ and $\mathrm{Var}\,Y = m/4$. Plugging this into the above, we get:
$$ E[Y(m - Y)] = \frac{m(m - 1)}{4} $$
Going back to our conditional expectation, we have then that:
$$ E[X_2 X_3 | X_1 = x_1 ] = \frac{(n - x_1)(n - x_1 - 1)}{4} $$
We can plug this into our sum above to write:
$$ E[ X_1 X_2 X_3 ] = \sum_{x_1 = 0}^{n} \frac{(n - x_1)(n - x_1 - 1)}{4} x_1 \binom{n}{x_1} \left(\frac{1}{3}\right)^{x_1} \left(\frac{2}{3}\right)^{n - x_1} $$
The last simplification we will make is we will note that the product $X_1 X_2 X_3$ is zero whenever any of the three is zero. This means that any terms with a zero will not contribute to the overall expectation. For this reason, we can ignore the cases when $x_1 \in \{0, 14, 15\}$ because any of those will guarantee at least one zero in the product. Keeping that in mind and substituting in for $n = 15$ our summation becomes:
$$ E[ X_1 X_2 X_3 ] = \sum_{x_1 = 1}^{13} \frac{(15 - x_1)(14 - x_1)}{4} x_1 \binom{15}{x_1} \left(\frac{1}{3}\right)^{x_1} \left(\frac{2}{3}\right)^{15 - x_1} $$
We can easily evaluate this sum on the computer to get an exact answer:
$$ E[ X_1 X_2 X_3 ] = \frac{910}{9} \approx 101.1 $$
So the expected product of the three dice roll frequencies aboue 101.
Disclaimer, for any generalizations the most reasonable approach would be to just compute the expectation directly on a computer -- the result is a closed form summation as I have below (although with upwards of $(N+1)^S$ terms.
Erdos
