Instead of taking three elements at a time and comparing them to every other three elements, which takes O(n2) comparisons, you can use the following, which only requires traversing the array once:
from collections import defaultdict
counts = defaultdict(int)
for i in xrange(len(string) - 2):
counts[tuple(string[i:i+3])] += 1
Using this answer about a Python equivalent to Ruby's each_cons, you can make this even nicer, though I don't know about the performance.
As the array contains characters, you may as well call it a string.