Why is [] faster than list()?

Because [] and {} are literal syntax. Python can create bytecode just to create the list or dictionary objects:

>>> import dis
>>> dis.dis(compile('[]', '', 'eval'))
  1           0 BUILD_LIST               0
              3 RETURN_VALUE        
>>> dis.dis(compile('{}', '', 'eval'))
  1           0 BUILD_MAP                0
              3 RETURN_VALUE        

list() and dict() are separate objects. Their names need to be resolved, the stack has to be involved to push the arguments, the frame has to be stored to retrieve later, and a call has to be made. That all takes more time.

For the empty case, that means you have at the very least a LOAD_NAME (which has to search through the global namespace as well as the __builtin__ module) followed by a CALL_FUNCTION, which has to preserve the current frame:

>>> dis.dis(compile('list()', '', 'eval'))
  1           0 LOAD_NAME                0 (list)
              3 CALL_FUNCTION            0
              6 RETURN_VALUE        
>>> dis.dis(compile('dict()', '', 'eval'))
  1           0 LOAD_NAME                0 (dict)
              3 CALL_FUNCTION            0
              6 RETURN_VALUE        

You can time the name lookup separately with timeit:

>>> import timeit
>>> timeit.timeit('list', number=10**7)
0.30749011039733887
>>> timeit.timeit('dict', number=10**7)
0.4215109348297119

The time discrepancy there is probably a dictionary hash collision. Subtract those times from the times for calling those objects, and compare the result against the times for using literals:

>>> timeit.timeit('[]', number=10**7)
0.30478692054748535
>>> timeit.timeit('{}', number=10**7)
0.31482696533203125
>>> timeit.timeit('list()', number=10**7)
0.9991960525512695
>>> timeit.timeit('dict()', number=10**7)
1.0200958251953125

So having to call the object takes an additional 1.00 - 0.31 - 0.30 == 0.39 seconds per 10 million calls.

You can avoid the global lookup cost by aliasing the global names as locals (using a timeit setup, everything you bind to a name is a local):

>>> timeit.timeit('_list', '_list = list', number=10**7)
0.1866450309753418
>>> timeit.timeit('_dict', '_dict = dict', number=10**7)
0.19016098976135254
>>> timeit.timeit('_list()', '_list = list', number=10**7)
0.841480016708374
>>> timeit.timeit('_dict()', '_dict = dict', number=10**7)
0.7233691215515137

but you never can overcome that CALL_FUNCTION cost.