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.