Assuming a 50-over match, the maximum number of deliveries a batsman can face is 49 * 5 + 6 = 251 (i.e., a batsman coming in as opener facing first ball of the innings can face all 6 balls of the first over and all except first ball of the remaining overs because strike will have to be rotated at end of over.) And theoretically, he can score a six off every fall he faces. (that’s 1506 runs, much more than the max. total number of runs ever scored in a one-day match…!)

The answer is 1653 runs.

Explanation:
For an ideal case scenario, the (opening) batsman will hit a six on each ball. But if he hits a six on the last ball of the over, the strike will no longer remain with him in the next over. Thus, the best thing he can do on the last ball is to run 3 runs so that he retains the strike with him even in the next over(s). Thus the total runs that he can score in each over:
(6 * 5) + 3 = 33 runs.
But this will have to continue like above only till the 49th over; this is the actual twist, as in the last over, he can hit a six on the last ball as well because that will be the last ball of the match (and there is no purpose of giving the strike to the other batsman before the over ends).
Thus runs for the last over will be  6 * 6 = 36 runs.
Therefore, the maximum runs will be  (33 * 49) + 36 = 1653 runs.  (Theoretically)

The answer is actually infinite.

This is because there is no limitation to the maximum number of runs scored in a single ball. A batsman can run for how much ever he could. Even if there were no overthrows, he might be capable of running 4 runs or even 5(think of him as Usain Bolt…), or the fielder may be too slow to run after the ball and retrieve it in time.

We can never be sure how much runs he could score. According to theory, like NitinGurbani has said, it is 1653 runs. But  we can’t be exactly sure.