recipients = list(itertools.chain.from_iterable(map(operator.attrgetter('to'), mail.outbox)))
recipients = [item for sublist in map(operator.attrgetter('to'), mail.outbox) for item in sublist]
recipients = sum(map(operator.attrgetter('to'), mail.outbox), [])
recipients = reduce(operator.add, map(operator.attrgetter('to'), mail.outbox))

[leaf for tree in forest for leaf in tree]

python -mtimeit -s'from itertools import chain; l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'list(chain.from_iterable(l))'
10000 loops, best of 3: 32.7 usec per loop
python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' '[item for sublist in l for item in sublist]'
10000 loops, best of 3: 60.8 usec per loop
python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'sum(l, [])'
1000 loops, best of 3: 477 usec per loop
python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'import operator' 'reduce(operator.add,l)'
1000 loops, best of 3: 498 usec per loop
python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'reduce(lambda x,y: x+y,l)'
1000 loops, best of 3: 514 usec per loop

Explanation: the shortcuts based on + (including the implied use in sum) are, of necessity, O(L**2) when there are L sublists -- as the intermediate result list keeps getting longer, at each step a new intermediate result list object gets allocated, and all the items in the previous intermediate result must be copied over (as well as a few new ones added at the end). So (for simplicity and without actual loss of generality) say you have L sublists of I items each: the first I items are copied back and forth L-1 times, the second I items L-2 times, and so on; total number of copies is I times the sum of x for x from 1 to L excluded, i.e., I * (L**2)/2.

The list comprehension just generates one list, once, and copies each item over (from its original place of residence to the result list) also exactly once.