Is there a better way to do this probability calculation? Suppose we consider the following probability problem: what is the probability of obtaining a total of $6$ or less when throwing $3$ dice?
My approach to this kind of problem is the following: the sample space is the following set
$$\Omega = \{(i,j,k) : 1\leq i,j,k\leq 6\}$$
Then we define a random variable $\xi : \Omega \to \mathbb{R}$ which gives the total amount of points, $\xi(i,j,k) = i+j+k$. In that case we want $P(\xi \leq 6)$. Now the event that corresponds to that is the event $A = \bigcup_{i=1}^6 A_i$ where $A_i$ is the event when the total score is $i$. So we have
$$P(\xi \leq 6) = P\left(\bigcup_{i=1}^6 A_i\right)=\sum_{i=1}^6 P(A_i)$$
Then we compute each $P(A_i)$ by in fact writing down $A_i$, and using the relation
$$P(A_i) = \dfrac{|A_i|}{|\Omega|}$$
Now, this seems a bad approach, because we have to list all the elements of $A_i$ for each $A_i$. On more complex situations this could not be nice to handle (in truth in this situation it is already quite tedious).
So, is there a better approach to tackle this problem?
 A: The event $A$ consists of $|A|$ equally-likely elements of the probability space $\Omega$,
where $|A|$ is the number of ways that we can form a sum of $6$ or less with three
ordered positive integers.
For example, $(1,1,2) \in A$ because $1 + 1 + 2 \leq 6$,
and $(2,1,1) \in A$ because $2 + 1 + 1 \leq 6$,
and moreover these are two distinct events.
The number $|A|$ is also the number of ways we can distribute $6$ or fewer
indistinguishable balls into three identified bins
(think of the number of balls in each bin as the number of pips showing on each die),
which in turn is the number of ways we can distribute exactly $7$ balls
into four identified bins (where the first three bins are identified with the
three bins in the previous model, and the fourth bin holds the balls that
were not put in bins in that model).
And we can count $|A|$ by putting one ball into each of the four bins
and then distributing the remaining $3$ balls in those bins, zero or more
of those balls in each bin.
So $|A|$ is the number of ways to distribute $3$ indistinguishable balls 
into $4$ identified bins, where there is no minimum number of balls to put in any bin.
This can be solved by the "stars and bars" method.
Once you have found $|A|$ by this method, you can set  $P(A) = \dfrac{|A|}{|\Omega|},$
using the same reasoning by which you proposed to find $P(A_i).$
Another possibility is to find each of your $A_i$ by distributing exactly $i-3$
balls into three bins with no minimum number of balls per bin,
and compute $P(A_i)$ as you have indicated.
This will come to the same result for fairly obvious reasons.
This method works for totals up to $8$ (just change the number of balls distributed
into the bins). If you want the probability of a total of $9$ or less, you have
to take into account the fact that $(1,1,7)$ (for example) is not in your
probability space; the simple stars-and-bars method alone does not account for that.
A: We assume that the results for each die are independent, and that the die is unbiased where each of the $6$ outcomes is equally likely. Denote as $S_n$ the sum resulting from the throw of $n$ dice.
For $S_1$ (the throw of one die), we can represent the probability mass function (pmf) as a vector $[1/6,1/6,1/6,1/6,1/6,1/6]$. or $\frac{1}{6}[1,1,1,1,1,1]$, where the index of each element corresponds to the number shown on the die in ascending order from $1$ to $6$ inclusive.
Consider the simple scenario of the sum of two dice. The minimum value will be $2$ and the maximum value will be $12$, leading to the pmf vector being of length $11$. Given the assumption of independence, the pmf of the sums can be obtained by the discrete convolution of two pmf (vectors) of length $6$, each containing $1$s, as follows:-
$$pmf(S_2)=\frac{1}{6}[1,1,1,1,1,1]\ast\frac{1}{6}[1,1,1,1,1,1]=\frac{1}{36}[1,2,3,4,5,6,5,4,3,2,1]$$
Note that sum of the vector is $1$, and that $S_2=7$ is the mode of the distribution, occurring at frequency $1/6$.
For $S_3$, the sum of three dice, we simply convolve $pmf(S_2)$ by vector $\frac{1}{6}[1,1,1,1,1,1]$, resulting in the length $16$ pmf vector  $pmf(S_3)$, where the first element corresponds to the minimum sum of $3$ and the last element corresponds to the maximum sum of $18$, as follows:-
$$pmf(S_3)=\frac{1}{216}[1,3,6,10,15,21,25,27,27,25,21,15,10,6,3,1]$$ Thus the number of throws that yield a total of $6$ or less will be $1+3+6+10=20$, resulting in a probability of $\frac{20}{216}$, or $\frac{5}{54}$.
By convolving pmf vector $\frac{1}{6}[1,1,1,1,1,1]$ with $pmf(S_n)$, we end up with the pmf of $S_{n+1}$ - performing this repeatedly for large $n$ leads to a pmf vector which tends to the Normal distribution, a visual demonstration of the Central Limit Theorem. 
A: It might be simpler to think about it in terms of [number of rolls less than or equal to 6]/[total number of possible rolls]
How many ways can we roll <= 6? First, how many ways can we roll less than six with two dice:
<= 2 - only 1 (1,1)
<= 3 - 3 (1,2),  (2,1),  (1,1)
<= 4 - 6 (1,2),  (2,1),  (1,1), (3,1), (1,3), (2,2)
For each step, we only need to add on pairs that equal that number exactly because the previous step contains all pairs less than the current number. You may also notice it follows the pattern of Trangular Numbers: 1, 3, 6, 10, 15, 21...
<= 5 - 10
Fitting in the last die is simple.
Suppose we roll a 4, from above we see there is one way to roll the other two dice to be less than or equal to 2.
Suppose we roll a 3, look above and see there are 3 ways for the other two dice to be less than or equal to 3. 
And so on. So add 1+3+6+10=20
There are 6^3 possible rolls. So 20/216 = 5/54 or about 9.259%.
A: More generally, Let $L(k,m)$ be the probability of obtaining a total $\le k$ when throwing $m$ fair dice.  Then (by conditioning on the result of the first die)
$$\eqalign{L(k,m) &= \dfrac{1}{6} \sum_{j=1}^6 L(k-j, m-1)\cr
L(k,m) &= 0 \ \text{if}\ k < m\cr
L(k,0) &= 1\ \text{for}\ k \ge 0}$$
Thus we get the following table:
$$ \matrix{ & k = 0 & 1 & 2 & 3 & 4 & 5 & 6\cr
m = 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1\cr
 1 & 0 & 1/6 & 2/6 & 3/6 & 4/6 & 5/6 & 6/6\cr
 2 & 0 & 0 & 1/36 & 3/36 & 6/36 & 10/36 & 15/36 \cr
 3 & 0 & 0 & 0 & 1/216 & 4/216 & 10/216 & 20/216 }$$
Alternatively, the generating function for $L(\cdot, m)$ is
$$g_m(x) = \dfrac{(x - x^7)^m}{6^m (1-x)^{m+1}}$$
