You should use a list comprehension:

>>> import pprint
>>> n = 3
>>> distance = [[[0 for k in xrange(n)] for j in xrange(n)] for i in xrange(n)]
>>> pprint.pprint(distance)
[[[0, 0, 0], [0, 0, 0], [0, 0, 0]],
 [[0, 0, 0], [0, 0, 0], [0, 0, 0]],
 [[0, 0, 0], [0, 0, 0], [0, 0, 0]]]
>>> distance[0][1]
[0, 0, 0]
>>> distance[0][1][2]
0

You could have produced a data structure with a statement that looked like the one you tried, but it would have had side effects since the inner lists are copy-by-reference:

>>> distance=[[[0]*n]*n]*n
>>> pprint.pprint(distance)
[[[0, 0, 0], [0, 0, 0], [0, 0, 0]],
 [[0, 0, 0], [0, 0, 0], [0, 0, 0]],
 [[0, 0, 0], [0, 0, 0], [0, 0, 0]]]
>>> distance[0][0][0] = 1
>>> pprint.pprint(distance)
[[[1, 0, 0], [1, 0, 0], [1, 0, 0]],
 [[1, 0, 0], [1, 0, 0], [1, 0, 0]],
 [[1, 0, 0], [1, 0, 0], [1, 0, 0]]]

The answers by @SonderingNarcissit and @MadPhysicist are already quite nice.

Here is a quick way of adding a number to each element in your list and keeping the structure. You can replace the function return_number by anything you like, if you want to not only add a number but do something else with it:

def return_number(my_number):
    return my_number + 4    

def add_number(my_list):

    if isinstance(my_list, (int, float)):
        return return_number(my_list)
    else:
        return [add_number(xi) for xi in my_list]

A = [[[0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0], [0], [0]]]

Then

print(add_number(A))

gives you the desired output:

[[[4, 4, 4], [4, 4, 4], [4, 4, 4]], [[4], [4], [4]]]

So what it does is that it look recursively through your list of lists and everytime it finds a number it adds the value 4; this should work for arbitrarily deep nested lists. That currently only works for numbers and lists; if you also have e.g. also dictionaries in your lists then you would have to add another if-clause.

newarray = np.dstack(mylist)

should work. For example:

import numpy as np

# Here is a list of five 10x10 arrays:
x = [np.random.random((10,10)) for _ in range(5)]

y = np.dstack(x)
print(y.shape)
# (10, 10, 5)

# To get the shape to be Nx10x10, you could  use rollaxis:
y = np.rollaxis(y,-1)
print(y.shape)
# (5, 10, 10)

np.dstack returns a new array. Thus, using np.dstack requires as much additional memory as the input arrays. If you are tight on memory, an alternative to np.dstack which requires less memory is to allocate space for the final array first, and then pour the input arrays into it one at a time. For example, if you had 58 arrays of shape (159459, 2380), then you could use

y = np.empty((159459, 2380, 58))
for i in range(58):
    # instantiate the input arrays one at a time
    x = np.random.random((159459, 2380))
    # copy x into y
    y[..., i] = x

If you build a 2D list and convert it to numpy with np.array() it will return a numpy array with 2 dimensions.

If you build a 3D list and convert it to numpy with np.array() it will return a numpy array with 3 dimensions.

Minimal working example:

my_list = [[[0,1,2],[0,1,2]],[[0,1,2],[0,1,2]]]
my_list

[[[0, 1, 2], [0, 1, 2]], [[0, 1, 2], [0, 1, 2]]]

my_np_array = np.array(my_list)
my_np_array.shape

(2, 2, 3)

I recommend using numpy for multidimensional arrays. It makes it much more convenient, and much faster. This would look like:

import numpy as np
x = np.zeros((10,20,30)) # Make a 10 by 20 by 30 array
x[0,0,0] = value1

Still, if you don't want to use numpy, or need non-rectangular multi-dimensional arrays, you will need to treat it as a list of lists of lists, and initialize each list:

x = []
x.append([])
x[0].append([])
x[0][0].append(value1)

Edit: Or you could use the compact notation shown in ndpu's answer (x = [[[value1]]]).

Using reshape and transpose:

a.reshape(-1, 2).T.reshape(-1, 3, 4)

array([[[ 0,  1,  2,  3],
        [ 4,  5,  6,  7],
        [ 8,  9, 10, 11]],

       [[12, 13, 14, 15],
        [16, 17, 18, 19],
        [20, 21, 22, 23]]])

Timings on your sample array:

%timeit np.rollaxis(a.reshape(3, 4, 2), 2)
2.92 µs ± 10.9 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%timeit a.reshape(2,4,3, order="F").swapaxes(1, 2)
1.1 µs ± 11.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%timeit a.reshape(-1, 2).T.reshape(-1, 3, 4)
1.08 µs ± 7.36 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

I haven't timed this answer on massive arrays because I haven't figured out a way to generalize either of your solutions. One benefit of my solution is that it scales without any altering of the code:

a = np.zeros(48)
a[::2] = np.arange(24)
a[1::2] = np.arange(24, 48) 
a.reshape(-1, 2).T.reshape(-1, 3, 4)

array([[[ 0.,  1.,  2.,  3.],
        [ 4.,  5.,  6.,  7.],
        [ 8.,  9., 10., 11.]],

       [[12., 13., 14., 15.],
        [16., 17., 18., 19.],
        [20., 21., 22., 23.]],

       [[24., 25., 26., 27.],
        [28., 29., 30., 31.],
        [32., 33., 34., 35.]],

       [[36., 37., 38., 39.],
        [40., 41., 42., 43.],
        [44., 45., 46., 47.]]])