I think you have a good hang of the concept. However, things can always be written better.
For example, the sample space for three independent dice, rather than the suggestive $\{(1,1,1),...,(6,6,6)\}$(which is correct, so credit for that) can be written succinctly as $\Xi \times \Xi \times \Xi$, where $\Xi = \{1,2,3,4,5,6\}$. This manages to express every element in the sample space crisply, since we know what elements of cartesian products look like.
The sigma field is a $\sigma$-algebra of subsets of $\Omega$. That is, it is a set of subsets of $\Omega$, which is closed under infinite union and complement.
Ideally, the sigma field corresponding to a probability space, is the set of events which can be "measured" relative to the experiment being performed i.e. it is possible to assign a probability to this event, with respect to the experiment being performed. What this specifically means, is that based on your experiment, your $\sigma$-field can possibly be a wise choice.
In our case, we have something nice : every subset of $\Omega$ can be "measured" since $\Omega$ is a finite set (it has $6^3 = 216$ elements) hence the obvious candidate, namely the cardinality of a set can serve as its measure(note : this choice is not unique! You can come up with many such $\mathcal F$).
Therefore, $\mathcal F = \mathcal P(\Omega)$, where $\mathcal P(S)$ for a given set $S$ is the power set of $S$, or the set of all subsets of $S$. This is logical since this contains every subset of $\Omega$, and is obviously a $\sigma$-algebra.
Now, $P(A)$, for $A \subset \mathcal F$(equivalently, $A$ any subset of $\Omega$) is, for the reasons I mentioned above, simply the ratio between its cardinality, and the cardinality of $\Omega$. Therefore, $P(A) = \frac{|A|}{216}$. That is exactly what you wrote, except well, division by sets isn't quite defined.
So there you have it, an answer, along with what you've done right and wrong.
NOTE : $\mathcal P$ for the power set, and $P$ for the probability of a set is my notation here. I still think the letter $P$ is used for both, but it's causing confusion here, hence the change.
Best Answer
You can use generating functions here.
Suppose we represent a single die with the function $x + x^2+x^3+x^4+x^5+x^6$
The different exponents here represent the different possible outcomes of rolling the die.
Now, if you want the different possible outcomes of rolling two dice, just multiply this function by itself:
$$(x + x^2+x^3+x^4+x^5+x^6)\cdot(x + x^2+x^3+x^4+x^5+x^6)$$
$$= x^2+2x^3+3x^4+4x^5+5x^6+6x^7+5x^8+4x^9+3x^{10}+2x^{11}+x^{12}$$
Again, the different exponents here represent the possible outcoems, so they range from $2$ to $12$
But what's really neat, is that the coefficients represent the number of ways to get that outcome.
For example, the term $5x^6$ indicates that there are $5$ ways to get an outcome of $6$
Why does this work? Well, we get one term $x^6$ by multiplying $x$ from the first term and $x^5$ from the second (i.e. by throwing a $1$ with the first die, and a $5$ with the second), but we also got a term $x^6$ by multiplying $x^2$ from the first term and $x^4$ from the second (i.e. by throwing a $2$ with the first die, and a $4$ with the second), etc.
OK, so for $6$ dice, we just need to find the different coefficients and exponents for the following function:
$$(x + x^2+x^3+x^4+x^5+x^6)^6$$
Here is where a tool like WolframAlpha comes in:
Scroll down to the expanded form, and you'll find:
$$x^{36} + 6 x^{3}5 + 21 x^{34} + 56 x^{33} + 126 x^{32} + 252 x^{31} + 456 x^{30} + 756 x^{29} + 1161 x^{28} + 1666 x^{27}$$
$$ + 2247 x^{26} + 2856 x^{25} + 3431 x^{24} + 3906 x^{23} + 4221 x^{22} + 4332 x^{21} + 4221 x^{20} + 3906 x^{19}$$
$$ + 3431 x^{18} + 2856 x^{17} + 2247 x^{16} + 1666 x^{15} + 1161 x^{14} + 756 x^{13} + 456 x^{12} + 252 x^{11} + 126 x^{10}$$
$$ + 56 x^9 + 21 x^8 + 6 x^7 + x^6$$
So, for example, given the term $2856x^{17}$ we know that there are $2856$ ways for the $6$ dice to add up to $17$, meaning that the probability of getting a sum of $17$ is $\frac{2856}{46656}$