You factor N! into N factors, and look at the smallest of those factors. t(N) is the largest that the smallest factor can be. Here are the best factorizations going slightly beyond the part of https://oeis.org/A034258 that was quoted by Terry Tao.

  1! =  1
  2! =  1 * 2
  3! =  1 * 2 * 3
  4! =  2 * 2 * 2 * 3
  5! =  2 * 2 * 2 * 3 * 5
  6! =  2 * 2 * 3 * 3 * 4 * 5
  7! =  2 * 2 * 3 * 3 * 4 * 5 * 7
  8! =  2 * 3 * 3 * 4 * 4 * 4 * 5 * 7
  9! =  3 * 3 * 3 * 4 * 4 * 4 * 5 * 6 * 7
  10! = 3 * 3 * 4 * 4 * 4 * 5 * 5 * 6 * 6 * 7
  11! = 3 * 3 * 4 * 4 * 4 * 5 * 5 * 6 * 6 * 7 * 11
  12! = 3 * 4 * 4 * 4 * 5 * 5 * 6 * 6 * 6 * 6 *  7 * 11
  13! = 3 * 4 * 4 * 4 * 5 * 5 * 6 * 6 * 6 * 6 *  7 * 11 * 13
  14! = 4 * 4 * 4 * 5 * 5 * 6 * 6 * 6 * 6 * 6 *  7 *  7 * 11 * 13
  15! = 4 * 5 * 5 * 5 * 6 * 6 * 6 * 6 * 6 * 6 *  7 *  7 *  8 * 11 * 13
  16! = 5 * 5 * 5 * 6 * 6 * 6 * 6 * 6 * 6 * 7 *  7 *  8 *  8 *  8 * 11 * 13

Is the number of factors when you write it out this way, always one more than the previous?

EDIT: Never mind, that's part of the definition of the problem!

If you don't need it to be fast, the naive way to implement "the largest quantity such that it is possible to factorize N! into N factors a_1, ..., a_N, each of which is at least t(N)" is

  import math
  factorize = lambda N, number_of_factors: [(factor,) + rest for factor in range(1, N+1) if N % factor == 0 for rest in factorize(N//factor, number_of_factors-1)] if number_of_factors > 1 else [(N,)]
  t = lambda N: max(min(factors) for factors in factorize(math.factorial(N), N))

The straightforward but slow way is to just brute-force over all possible factorizations of N!.

As the post states, writing N! as a product of factors is equivalent to writing log(N!) as a sum of those factors' logarithms. So log(t(N)) is the smallest bin size such that the factors all "fit" into N bins of that size.

This computation is simple to describe and implement, it's just inefficient because there's a combinatorial explosion of possible ways to pack factors into bins. It's an instance of the knapsack problem which is NP-complete.

It's a subset of the knapsack problem though, so there may well be some shortcut to solving it that doesn't let us solve arbitrary knapsacks.

Right, I was just trying to clarify because the OP was asking for a way to compute the function. It may not be the best way, and that's the interesting open question.