Sum of three dice is eleven the sample space of one toss of three dice is:
$\Omega = \left \{ (1,1,1), ..., (6,6,6) \right \}$
so there are $6^3 = 216$ possible outcomes.
What is the probability to obtain an outcome where the sum of its three components is equal to 11?
I've considered the possible value can assume dice without a particular position and then I have considered the permutations to include every position:
$(6,4,1), 3! = 6 \\ (6,3,2), 3! = 6 \\ (5,5,1), \frac{3!}{2!} = 3 \\ (5,4,2), 3! = 6 \\ (5,3,3), \frac{3!}{2!} = 3 \\ (4,4,3), \frac{3!}{2!} = 3$
so I've summed up obtaining $27$ and the probability would be $\frac{27}{216}$
Now, consider if I have to do this same passages for sums from 3 to 18, it is very exhausting.
So, my question is: Is there any "faster" way to do that?

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 $n = 11$.
