How can I combine several ranked lists into one mega list?

This is a more subtle question than you might expect. It's equivalent to asking the "best" way to rank candidates in an election, but Arrow's impossibility theorem shows that there really is no "best" way to do this (more precisely, it is impossible to prevent irrelevant candidates from spoiling the results).

So the best we can do is acknowledge the drawbacks of whatever algorithm we decide to use.

The most straightforward approach would be to use the Borda method. In this approach, we first tabulate the N unique candidates from all the ballots (lists). Then for each ballot, we compute a Borda score for each candidate in that ballot, where the Borda score is given by

Borda score = N - rank

If a candidate is unranked in a given ballot, then its rank is assumed to be N.

For example, given two ballots, we would have the following Borda scores

which would produce the following ranked list (by summing the Borda scores for each candidate)

Unfortunately, the Borda method gives poorer predictions when some ballots don't rank every candidate, as a ballot can spoil the final result by ranking their favorite candidate first and leaving the other candidates unranked. For example, imagine you want to combine two video game lists. One list was compiled by an IGN critic who meticulously tabulated their hundred favorite games on the Nintendo DS. The other ballot was written by a 7-year-old and contains only My Little Pony: Pinkie Pie's Party. Then the Borda method would rank My Little Pony: Pinkie Pie's Party above every game in the critic's list except the critic's top game, which would be rated equal.

Since lists of favorite media are unlikely to have complete overlap, you find yourself vulnerable to this shortcoming. However, we can attempt to correct for this failure mode! In this paper, the authors consider some other formulations of the Borda method. The simplest alternative to implement would be the average Borda count method.

The average Borda count method is equivalent to the Borda method except when a ballot contains unranked candidates. In this case, the unranked candidates on the Ballot receive a score equal to the average of the unused points that the ballot could have allotted. With a little bit of algebra, you can show that the average Borda count is given by

ABC = (N - K - 1) / 2

where K is the number of items on a given ballot. For the example from before, all but one point was allotted, so the average Borda count for the first list would be (5-3-1)/2=0.5 whereas the second list would not have an average Borda count since all the candidates are ranked. Now we have

which becomes

In this case, there is no change in the final ordering. But in the critic vs 7-year-old example, MLP would be ranked in the middle of the list instead of at the front (for better or worse).

There are a few tools available online to compute the Borda scores for you. Here is one such calculator, though it only allows for five candidates (almost certainly too few for your use case). A few people have also written python scripts, for example here. Unfortunately, neither of these implementations correct for the average Borda count, so you would have to do that yourself.