The value under the optimal subsequent strategy of testing a new lever is a least as good as any particular strategy that tests a new lever on the next step. We will consider the particular strategy of testing new levers until we find 1274. To simplify the calculation, we will only consider profits from the 1274 lever as a lower bound on the value. With $n$ unknown levers and $5+n$ days remaining (if we have tested every day, the only case we will be concerned with), the expected number of pulls of the 1274 lever under this strategy is $((5+n) + (4+n) + (3+n) + \dots + 7 + 6)/n=(n+11)/2$ for an expected daily value of $1274(n+11)/(2n+10)$. The value is less when $n$ is larger, so in the worst case of $n=9$, we still expect to earn at least 910 per day, more than the second best lever.
Any strategy that involves using a known lever before testing a new lever is equivalent to using the known lever at the end instead (it can be delayed since it provides no information). (Actually it is obviously non-optimal unless we have found 1274 already since at the end we should use the best known lever then which might be better than the best we have found now.) So, we only need to consider strategies that do all their testing before pulling any known levers. Among such strategies, the best strategy that doesn't test on the next step is obviously just pulling the best known lever on all remaining days which can't do better than 728 per day unless we have found 1274. (If we have found 1274, obviously pulling that is best.)
Thus, we can't do better (in expectation) than testing until we find 1274.
If we care to calculate the exact expected value of this strategy, we take the expected value from the 1274 lever which is $14 \cdot 910$ and add half the value of every other unknown lever (since there is a 50% chance to find each particular lever before 1274; and linearity of expectation) to get 13401.5.