Not base 2. All palindromes in base 2 have a 1 in the last position, because there's a 1 in the first position. This means that any addition of 3 palindromes must be odd because the result of the addition is guaranteed a 1 in the ones position, meaning even numbers can't possibly be represented that way. This is discussed in the original paper at the bottom of the second page: https://arxiv.org/pdf/1602.06208.pdf , which I've expanded out a bit since they just mention the evenness without the argument I give. They say base 2 needs at least 4 summands, but their phrasing suggests to me that's intended as a lower bound, not that they have a proof 4 is always sufficient. (Not to mention it would have to be more complicated than that anyhow since it would have to be something like "3 for odd and 4 for even above some threshold", or some other bounds.)
If you pause at 3:33 in the video and read the abstract, it actually says that numbers in base 2 require "at most 4 palindromes", rather than exactly 3, and "similar results" were found for bases 3 and 4. So it's not quite the same result as for bases of 5 or greater.
