Here is a rough estimate:
Suppose that we never roll 2,3,11,12,or 7. How many rolls do we need to collect the six possible values? If the six outcomes were equally likely, the expected number of rolls we need would be 6/6+6/5+6/4+6/3+6/2+6/1 which is about 15. The true probabilities are (4,5,6,6,5,4)/24, so the uniform approximation shouldn't be too bad.
Now assume we never roll 2,3,11,12. We need to avoid 7 about 15 times to win. The probability of this is (24/30)15 which is about 1/27.
Finally, rolling 2,3,11,or 12 doesn't affect the outcome of the games, so we get the same success probability when these rolls are possible.
The true probability is somewhat higher because the function f(n) = (4/5)n is convex, and I evaluated f at the mean rather than taking the expectation over the distribution of hitting times.
To get a better estimate, you should do what others have suggested and simulate. There is a method to calculate the probability exactly, but it will be a bit tedious.