Sum of three dice is eleven
The generating function approach is to ask for the coefficients of $x^{n}$ in the expansion of:
$$(x+x^2+x^3+x^4+x^5+x^6)^{3}=x^3\left(\frac{1-x^6}{1-x}\right)^3$$
Now, $$\frac{1}{(1-x)^3}=\sum_{k=0}^{\infty}\binom{k+2}{2}x^k$$
And $$(1-x^6)^3=1-3x^6+3x^{12}-x^{18}$$
So the product of these and $x^3$ has, for the coefficient $x^n$:
$$\binom{n-1}{2}-3\binom{n-7}{2}+3\binom{n-13}{2}-\binom{n-19}{2}$$
Trick is to treat $\binom{j}{2}=0$ when $j<2.$
So, for example, when $n=11$, you get $\binom{10}{2}-3\binom{4}{2}=45-18=27$.
This also gives you a hint of the value when dealing with $k$ dice. Then:
$$c_n = \sum_{i=0}^{k}(-1)^i\binom{k}{i}\binom{n-(1+6i)}{k-1}$$
If you don't like generating functions, this can be proven via inclusion-exclusion.
There's another approach that is a little faster for computing all values.
If $(1+x+x^2+\cdots+x^5)^3=a_0+a_1x+\cdots+a_nx^n+\cdots$ then you get that:
$$(a_0+a_1x+\cdots+a_nx^n+\cdots)(1-3x+3x^2-x^3)=(1-x^6)^3=1-3x^6+3x^{12}-x^{18}$$
What this means is that (setting $a_n=0$ when $n<0$: $$a_{n}-3a_{n-1}+3a_{n-2}-a_{n-3}=\begin{cases} 1&n=0\\ -3&n=6\\ 3&n=12\\ -1&n=18\\ 0&\text{otherwise} \end{cases}$$
or:
$$a_{n}=3\left(a_{n-1}-a_{n-2}\right)+a_{n-3}+\begin{cases} (-1)^k\binom{3}{k}&n=6k\\ 0&\text{otherwise} \end{cases}$$
Then the final coefficient of $x^n$, after we multiply by $x^3$ again, is $a_{n-3}$.
So you get:
$$\begin{align}a_0&=0+(-1)^{0}\binom{3}{0}=1\\ a_1&=3\left( a_0-a_{-1}\right)=3\\ a_2&=3\left( a_1 - a_0\right)=6\\ a_3&=3\left(a_2 - a_1\right)+a_0=10\\ a_4&=3\left(a_3-a_2\right)+a_1=15\\ a_5&=3\left(a_4-a_3\right)+a_2=21\\ a_6&=3\left(a_5-a_4\right)+a_3+(-1)^{1}\binom{3}{1}=25\\ &\cdots \end{align}$$
There's a special trick that can be applied here: $a_{n}=3a_{n-1}-3a_{n-2}+a_{n-3}$ is known to be a quadratic polynomial in $n$. So, when $n$ is not a multiple of $6$, you get that:
$$a_{n-2}-a_{n-3},a_{n-1}-a_{n-2},a_{n}-a_{n-1}$$ must be an arithmetic progression for any $n$ not a multiple of $6$.
So we have that $a_5-a_4 = 6, a_6-a_5=4$ and thus $a_7-a_6=2$, or $a_7=a_6+2=27.$ $a_8=27+0=27, a_9=27-2=25,a_{10}=25-4=21,a_{11}=21-6=15,a_{12}=15-8+(-1)^2\binom{3}{2}=10$.
For the favorable cases: Since the dice can assume integer values between $1$ and $6$, we wish to find the number of solutions in the positive integers of the equation $$x_1 + x_2 + x_3 = 11 \tag{1}$$ subject to the restrictions that $x_k \leq 6$ for $1 \leq k \le 3$.
A particular solution of equation 1 corresponds to the placement of two addition signs in the ten spaces between successive ones in a row of eleven ones. For instance, $$1 1 1 1 1 + 1 1 + 1 1 1 1$$ corresponds to the solution $x_1 = 5$, $x_2 = 2$, and $x_3 = 4$. Hence, the number of solutions of equation 1 in the positive integers is the number of ways we can select two of the ten spaces between successive ones in a row of eleven ones in which to place addition signs, which is $$\binom{10}{2}$$
More generally, the equation $$x_1 + x_2 + \cdots + x_k = n$$ has $$\binom{n - 1}{k - 1}$$ solutions in the positive integers since we must choose which $k - 1$ of the $n - 1$ spaces between successive ones in a row of $n$ ones will be filled with addition signs.
However, we have counted solutions in which one of the variables exceeds $6$. Notice that it is not possible for two of the variables to exceed $6$ simultaneously since $2 \cdot 7 = 14 > 11$.
Suppose $x_1 > 6$. Let $x_1' = x_1 - 6$. Then $x_1'$ is a positive integer. Substituting $x_1' + 6$ for $x_1$ in equation 1 yields \begin{align*} x_1' + 6 + x_2 + x_3 & = 11\\ x_1' + x_2 + x_3 & = 5 \tag{2} \end{align*} Equation 2 is an equation in the positive integers with $$\binom{5 - 1}{3 - 1} = \binom{4}{2}$$ solutions. By symmetry, there are an equal number of solutions in which $x_2 > 6$ or $x_3 > 6$. Hence, the number of solutions we must exclude is $$\binom{3}{1}\binom{4}{2}$$
Hence, the number of permissible solutions is $$\binom{10}{2} - \binom{3}{1}\binom{4}{2}$$
Addendum: You asked about the sums from $3$ to $18$. If $3 \leq n \leq 8$, then the number of solutions of the equation
$$x_1 + x_2 + x_3 = n \tag{3}$$
in the positive integers with $x_k \leq 6$ for $1 \leq k \leq 3$ is
$$\binom{n - 1}{3 - 1} = \binom{n - 1}{2}$$
since it is not possible for one of the variables to exceed $6$.
Notice also that by symmetry, the number of solutions with sum $3$ (all ones) is equal to the number of solutions with sum $18$ (all sixes), the number of solutions with sum $4$ (two ones and a two) is equal to the number of solutions with sum $17$ (two fives and a six), and so forth. Hence, knowing the number of solutions for $3 \leq n \leq 8$ also tells us the number of solutions for $13 \leq n \leq 18$.
The following argument shows that the number of solutions with sum $n$ is equal to the number of solutions with sum $21 - n$. If $y_k = 7 - x_k$, then $1 \leq x_k \leq 6 \implies 1 \leq y_k \leq 6$. Moreover, substituting $7 - y_k$ for $x_k$ for $1 \leq k \leq 3$ in equation 3 yields \begin{align*} 7 - y_1 + 7 - y_2 + 7 - y_3 & = n\\ 21 - y_1 - y_2 - y_3 & = n\\ -21 + y_1 + y_2 + y_3 & = -n\\ y_1 + y_2 + y_3 & = 21 - n \end{align*}
By an argument similar to that given above for $n = 11$, the number of solutions of the equation $$x_1 + x_2 + x_3 = 9$$ in positive integers not exceeding six is $$\binom{8}{2} - \binom{3}{1}\binom{2}{2} = \binom{8}{2} - \binom{3}{1}$$ By symmetry, this is also the number of solutions for $n = 12$.
Also, by symmetry, the number of solutions for $n = 10$ is equal to the number of solutions for $n = 11$.
It is the coefficient of $x^{11}$ in $(x(1+x+x^2+x^3+x^4+x^5))^3$ which is indeed $\color{red}{27}$ but any which a ways you solve it will be a bit of a slog.
