The probability it stops when you have N pixies .7^N. As .7 is very close to 1/sqrt(2), this means the probability of stopping is very close to 1 in 2^(N/2).
The probability of it ever stopping after pixie N is thus basically equal to the sum of i from N+1 to infinity of 1/2^(i/2). As 1/2^((i+1)/2) is less than 1/2^(i/2), we can bound this above by the sum from (N+1)/2 to infinity of 2 * 1/2^i, which is simply 4 * 1/2^((N+1)/2), which is less than 4 / 2^(N/2).
So after 20 rolls, the odds you will ever fail are less than 0.4%.
This is depressingly slow. I suspect most players won't concede if their chance of surviving is over 1%. It is merely an upper bound; the actual value is going to be a touch lower, but not that much (under 25% difference, if my napkin math is correct).
The point where your chances of ever failing are under 1 in a million is after 40 pixies. The point where your chances of ever failing are under 1 in a trillion are after 80 pixies. The point where your chances of ever failing are under 1 in a googol is after 667 pixies.
Your opponent has probably conceded before you have 667 pixies. But you'd probably stop making pixies before you hit 667 as well.
The point, in number of Pixies, where you would reasonably be able to state "I am almost certainly able to make as many Pixies as I want" is almost always going to be at a point after you have enough Pixies to win.
There is no rule in MtG that translates "you are almost certainly going to be able to do X" into "you can do X". However, in this case it is pretty much moot, because the rate at which certainty grows isn't really fast enough to matter much.
It is reasonable for the rules to allow your opponent to ask you to roll 13d20 and ensure you get a 15+ if you want another pixie, as that has a 1% chance of failure.
At 20 pixies, getting at 15+ roll has a 0.08% chance of failure, and the accumulated chance of a failure at any point in the future is more like 0.25%; a 1 in 400 chance.
MtG doesn't say "this is almost certain, you win" at any point, and even if it does it would probably be well after 20 pixies.
It is, however, possible that if the defending player has Graham's number life and insisted that you continue to roll d20s to spawn more pixies after you have 100 of them. If this ever happens, I suspect the judge's decision will be scrutinized heavily.
"Unsporting conduct -- Stalling" may be appropriate here. The player with the infinite pixie combination is almost certainly going to win, and the decision not to concede and require the (near infinite) dice to be rolled (with a near zero chance of failure) is almost entirely aimed at running the round time out. This is going to be a strange case, because the player doing the actions is not the one possibly stalling. "Stalling" is about intent to use the time constraint, not actions, and the intent of requiring all the dice to be rolled is "I am going to lose, but I tie if I make all the dice be rolled due to time constraints".
A fair way to handle it
It isn't hard to calculate the cumulative chance of failure after having X pixies at any point from there to infinity. Pre-calculating that would provide a "fair" way to handle the Graham's (or other insane life totals; 100 and Graham's number chance of failing is not practically different) cases.
If F_i is the chance of failure when rolling id20 for a 15+, then an upper bound on the chance of ever getting a failure after X pixies is sum i from X to infinity of F_i.
(This is actually the expected total number of failures, which is higher than the chance of ever failing).
F_i is .7^i, the sum is .7^(X) + .7^(X+1) + ..., which is .7^X * (1 + .7^2 + .7^3 + ...) which is .7^X * (1/1-.7), which is 3.33 * .7^X.
Which in turn is less than rolling Xd20 4 times and passing all 4 tries.
The actual chance of any failure is:
.7^X + (1-.7^X).7^(X+1) + (1-(1-.7^X).7^(X+1)) * .7^(X+2) + ...
When .7^X is almost 0, this is very close to the above calculation (it is off by a factor of less than (1-.7^X)). 3.333/4 < (1-.7^X) when X>=5; so this approximation is fair (gives an advantage to your opponent) if you have at least 5 pixies already.
So at any point in this combo, when you have X (5 or more) pixies, asking "I want to continue until I have a near infinite number of pixies. The chance of failure is less than the chance of rolling Xd20 4 times and failing on any of the 4 rolls. How about I just do that instead?"
That gives the defender a greater chance of you failing than doing it manually, and it takes far less time. You, personally, choose to manually roll up to X pixies, where X is large enough that you feel comfortable with the advantage you are giving your opponent by not rolling forever.
(4 rolls seems unfair to say "I get to win an infinite number of rolls", but it works because each roll you win would make the chance of a later failure drop further; the improvement is so rapid that the 4 rolls without improvement is a fair handicap to match up against infinite rolls with improvement.)