Evaluate a minimax tree

Python 2, 53 bytes

f=lambda l,c=1:c*l if l<[]else-min(f(x,-c)for x in l)

The main question is how to alternate between max and min each layer. Using the fact that max(l) == -min([-x for x in l]), we instead negate every second layer and recurse with -min. To negate every second layer, we pass down a multiplier c that alternates +1 and -1, and when we reach a leaf, we adjust its value by the multiplier. We recognize being at a leaf via the condition l<[], since Python 2 treats numbers as less than lists.

It's hard to shorten the else/if ternary because either branch could give a Truthy (nonzero) or Falsey (zero) value.

Jelly, 7 bytes

N߀¬¡ṂN

Try it online! or verify all test cases.

This uses the algorithm from @xnor's answer. For comparison purposes, a more straightforward approach that alternately computes minima and maxima weighs 8 bytes:

߀¬¡€Ṃ€Ṁ

How it works

N߀¬¡ṂN  Main link. Argument: x (array or integer)

N        Negate. For an array, this negates all integers in it.
   ¬     Logical NOT. For an array, this applies to all integers in it.
    ¡    Apply the second link to the left once if ¬ returned an array or 1, or not
         at all if it returned 0.
 ߀      Recursively call the main link on all elements at depth 1 of -x.
         If -x == 0, € will cast 0 to range before mapping ß over it. 
         Since the range of 0 is [], mapping ß over it simply yields [].
     Ṃ   Minimum.
         For an integer, Ṃ simply returns the integer. For [], Ṃ returns 0.
      N  Negate the minimum.