If you are willing to sample literally forever it seems like you'd be able to convince the Bayesian that the mean is not 0 with arbitrary Bayes factor.

No, there's a limit on that as well. See http://www.ejwagenmakers.com/2007/StoppingRuleAppendix.pdf

Instead, as pointed out by Edwards et al. (1963, p. 239): “(...) if you set out to collect data until your posterior probability for a hypothesis which unknown to you is true has been reduced to .01, then 99 times out of 100 you will never make it, no matter how many data you, or your children after you, may collect (...)”.

If you can generate arbitrarily high Bayes factors, then you can reduce your posterior to .01, which means that it can only happen 1 in 100 times. You can never have a guarantee of always getting strong evidence for a false hypothesis. If you find a case that does, it will be new to me and probably change my mind.

But it doesn't change the results if we switch and say we fool 33% of the Bayesians with Bayes factor of 3. We are still fooling them.

That doesn't concern me. I'm not going to argue for why, I'll just point out that if it is a problem, it has absolutely nothing to do with optional stopping. The exact same behavior (probability 1/3 of generating a Bayes factor of 3 in favor of a false hypothesis) shows up in the following case: a coin either always lands on heads, or lands on heads 1/3 of the time and tails 2/3 of the time. I flip the coin a single time. Let's say the coin is the second coin. There's a 33% chance of getting heads, which would produce a Bayes factor of 3 in favor of the 100%H coin.

If there's something wrong with that, it's a problem with classic Bayes, not optional stopping.

It is my thesis that every optional stopping so-called paradox can be converted into a form without optional stopping, and those will be clearer as to whether the problem is real or not.