A quick answer is the value of the maximum you're allowed to bet. Just wait for a 22 and then play it, resulting in walking away from the table with an extra unit.
A more rigorous answer should look at the cost of waiting for that 22. This is something I don't know exactly, but can estimate.
The cost to the player of the dealer-push-22 rule is 6.91%. However, the dealer will get to 22 more than that, because the player won't care if he busted first. Depending on the rules, the player will bust about 15.7% of the time. If we assume this is uncorrelated with the dealer's probability of busting (which it's not), then the probability of a dealer bust is 6.91%/(1.0-0.157) = 8.2%.
I know this is rough, but let's assume the player's 22 probability is the same. I know the dealer will bust more than the player, but this coupon also doesn't count after a split, so let's just say those factors offset. So, if the player gets a 22 8.2% of hands, then he will get a 22 once every 1/0.082 = 12.2 hands on average.
Let's say the house edge in the base game is 0.75%. The cost of playing 12.2 hands at that house edge is 12.2*0.0075 = 0.0915. So, subtract that from the one unit value and you get a value of the coupon of 1.0 - 0.0915 about 91% of face value.
This question is raised and discussed in my forum at Wizard of Vegas.