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
Let $q(x)=\frac{1}{6}\left(x^1+x^2+x^3+x^4+x^5+x^6\right)$. Write:
$$\begin{align}G(x)&=\sum_{n=0}^\infty q(x)^n\\&=\frac{1}{1-q(x)}\\&=\frac{6}{6-x-x^2-x^3-x^4-x^5-x^6} \end{align}$$
Then the coefficient of $x^N$ in $G(x)$ is your probability.
So you need to know something about the roots of the denominator. $1$ is one such root, but you are going to need to deal with complex roots, as well, or at least get an upper bound for the roots.
The formula is:
$$p(N)=a_1\alpha_1^N+a_2\alpha_2^N+\cdots a_6\alpha_6^N$$
For some constants $a_i$ and with the $\alpha_i$ the inverse roots of the denominator.
According to Wolfram Alpha, the roots of the denominator are $1$ and a bunch of values (one real, four complex) with absolute value greater than $1$.
So for large enough $N$, $p(N)\approx a_1$.
We can find $a_1$ by doing the usual partial fractions computation:
$$\lim_{x\to 1} G(x)(1-x)=\frac{2}{7}=\frac{1}{3.5}$$ which is what Arthur conjectured in comments above.
It turns out for $i>1$, $|\alpha_i|<\frac{3}{4}$, so we have:
$$p(N)=\frac{2}{7} + o\left(\left(\frac 34\right)^N\right)$$
We can also get a recursion:
$$P(n+6)=\frac{1}{6}\left(P(n)+P(n+1)+P(n+2)+P(n+3)+P(n+4)+P(n+5)\right)$$
We get $P(28)=.2857\dots.$ For $n\geq 88$ we get:
$$P(n)=0.285714285714\dots$$ which is $12$ digits of accuracy to $\frac{2}{7}=0.\overline{285714}.$
We also get that $$\frac{P(100)-\frac{2}{7}}{\left(\frac34\right)^{100}}\approx -0.016$$
and
$$\frac{P(1000)-\frac{2}{7}}{\left(\frac34\right)^{1000}}\approx 7.34\times 10^{-13}$$
(Computing the left side with python Fraction class to get exact value, then converted to float.)
This gives:
$$P(1000)-\frac{2}{7}\approx 8.45\times 10^{-138}$$