As usual for this type of problem, we should work on the smaller problem and try find a general formula or a way to use a computer to do it instead. I'll assume your calcs are correct in that only consecutive double crits will kill.
For remaining PP = 1, the probability of a double crit is 0 trivially.
For remaining PP = 2, the probability of a double crit is 1/256 ~ 0.4% as calculated above.
For remaining PP = 3, the probability is Pr(double crit in 1st 2) + Pr(1st hit no crit & next 2 are) = 1/256 + 15/16 * 1/256, ~0.0075683594 or 0.75%.
Remaining PP = 4, Pr(double crit in 1st 3) + Pr(double crit ONLY in last 2) + Pr(1st is crit, 2nd no crit, then double crits) = 0.75% as above + 15/16 * 15/16 *1/256 + 1/16 * 15/16 * 1/256
And so forth -- this is an easy thing for a computer to do but tedious by hand.
Your quick estimate for PP = 3 would be
Pr(next 3 don't double crit) = 0.99609375^2
Pr(double crit) = 1 - above = 0.0077972412
So yours is an overestimate (more pessimistic than reality). It will obviously get worse as PP increases, but how much worse?
Not too bad actually!
Using
https://statisticsfundamentals.com/calculators/coin-toss-streak-calculator/
We see that the probability is 5.06% compared to your 5.33%.
Edit to add: I also verified my calculation for PP = 3 and PP = 32 (for HP Fire) has a double crit probability a touch over 10%.