In answering this type of question, it is often useful to answer the opposite question!
"At least..." immediately suggests multiple routes, with multiple calculations, while "never" is more easily dealt with
So, there is a probability of 22/36 of not succeeding on the first roll of two dice, and again the same probability of not succeeding on the second roll. The probability of failing is thus (22/36)^2. The probability of succeeding is then (1 - 484/1296) or around 0.6265
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
The trick of using base-$6$ expansions on $[0, 1]$ in the comments is nice but you might find it a little inelegant, since after all the problem statement makes no reference to real numbers, and there is this non-uniqueness issue (it doesn't matter since it does not affect the probability of anything - we only have "non-uniqueness up to a set of measure zero" - but I think it's a little conceptually unsatisfying).
Alternatively there is a general construction of the countable product of probability measures, which is relatively simple in this case. The measurable sets are taken to be the cylinder $\sigma$-algebra. This is just saying the measurable sets are generated by the events "the first $n$ dice rolls have these exact values." The probability measure is determined by what it does to these events and those probabilities should be the obvious thing (specifying the first $n$ dice rolls gives an event of probability $\frac{1}{6^n}$). This measure can be thought of as Haar measure on the product of infinitely many copies of the finite cyclic group $\mathbb{Z}/6$.
If you believe that this all works out then the desired statement follows by the second Borel-Cantelli lemma.