How do I create a nested dictionary under a key that is yet to exist?

You could use collections.defaultdict, passing the default factory as dict:

>>> from collections import defaultdict
>>> d = defaultdict(dict)
>>> d['key']['subkey'] = 'value'
>>> d
defaultdict(<type 'dict'>, {'key': {'subkey': 'value'}})

To apply further levels of nesting, you can create a defaultdict that returns defaultdicts to a n-th depth of nesting, using a function, preferably anonymous, to return the nested default dict(s):

>>> d = defaultdict(lambda: defaultdict(dict))
>>> d['key']['subkey']['subsubkey'] = 'value'
>>> d
defaultdict(<function <lambda> at 0x104082398>, {'key': defaultdict(<type 'dict'>, {'subkey': {'subsubkey': 'value'}})})

_{Example shows nesting up to depth n=1}

You are using a [] list literal not a {} dict literal:

array['key'] = {}
array['key']['subkey'] = 'value'

But this isn't very useful in a loop.
In a loop you could test if 'key' is not in array - which is a cheap operation (O(1) lookup):

if 'key' not in array:
    array['key'] = {}
array['key']['subkey'] = 'value'

But you can use setdefault() to do the same thing and give key a default value if it doesn't already have a value, e.g.:

array.setdefault('key', {})['subkey'] = 'value'

And if this looks ugly, then you can always use collection.defaultdict.