In a rudimentary manner, favorable ways $=\dbinom{20}{x}\cdot 1^x\cdot 11^{20-x}$, against $12^{20}$ total ways,
or you could use the familiar binomial distribution,
$P(X = x) = \dbinom{20}{x}\cdot\left(\dfrac{1}{12}\right)^x\cdot\left(\dfrac{11}{12}\right)^{20-x}$
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
Did you also get these probabilities:
I did it with a Markov chain where the states are sorted 3-tuples of how many we have gotten of each value and there is an absorbing state "we have gotten at least 10 of some value". In another words the states at level $k$ (after $k$ dice throws) are the integer partitions of $k$ into at most $3$ parts. Note: we don't have to keep track of "which is which" in the counts tuple, always sort it when checking the transitions.
Here's a picture of the transition graph for $n=4$ (the direction is towards larger sum (and always towards the absorbing state)):
There were $221$ states in total for the $n=10$ case. Note: the matrix is very sparse (only max $3$ transitions from each state), so this can quicken the calculations of its powers (which we now want to calculate all upto $28$ so iteratively it is :-)).
I noticed that for $m=10, 11, \dots, 19$ the formula
$$P(X=m) = \frac{ {{m-1}\choose{9}}2^{m-10} }{3^{m-1}}$$
works and for $m=20$ we subtract the central binomial coefficient:
$$P(X=20) = \frac{ {{m-1}\choose{9}}2^{m-10} - {{20}\choose{10}} }{3^{m-1}}$$
but after that I don't see how the formula should be modified. And I don't immeaditely see a reason for that formula either, I just found it by checking OEIS with the numerators of the probabilities.
EDIT Then, for $m=21$:
$$P(X=21) = \frac{ {{m-1}\choose{9}}2^{m-10} - {{m}\choose{10}}2 - {{m/2}\choose{7}}2^{14} }{3^{m-1}}$$
Notice, there is a generilized binomial coeffient ${{21/2}\choose{7}} = \frac{415701}{2048}$ in there.
My guess is it comes some how straigh-forwardly from the matrix multiplication (there are punch of $1$'s and $2$'s in there and of course the denominator $3$) upto some point, but when the matrix becomes "full enough", the pattern breaks.
EDIT The matrix is constructed by first finding all the possible states and initializing it to a zero matrix. Then, for each state (it is a triplet) find the transitions (total $3$ of them) by adding a $1$ to a particular place and sorting the outcome. Then find the index of that in the states list (or if it goes over $n$ take the last index, which represents the absorbing state) and increment the matrix by $1$ in the correct spot. So we keep the matrix as an integer matrix but the probabilities are obtained by dividing by $3$ (and by $3^t$ in the $t$th power).
Here's the Python code (It's kind of messy, since I changed the matrix multiplication calculations to be done by only storing the non-zero entries of the matrix to speed up the calculation. But the methods for making the states and the matrix are first):