Python 2, 62 61 bytes

Zero-based. I assume the L suffixes are acceptable.

lambda n:[5**int(n/.7-~i)/10**n%10for i in range(2**n/2)]or 5

Output:

0 [5]
1 [2]
2 [1, 6]
3 [3, 5, 8, 0]
4 [1, 7, 9, 5, 6, 2, 4, 0]
5 [3, 9, 7, 8, 1, 7, 5, 5, 8, 4, 2, 3, 6, 2, 0, 0]
6 [1, 9, 8, 4, 0, 3, 7, 7, 9, 7, 6, 1, 8, 1, 5, 5, 6, 4, 3, 9L, 5L, 8L, 2L, 2L, 4L, 2L, 1L, 6L, 3L, 6L, 0L, 0L]
7 [4, 4, 2, 0, 1, 8, 3, 9, 8, 3, 5, 9, 5, 7, 7, 8, 2, 1L, 9L, 7L, 9L, 6L, 1L, 7L, 6L, 0L, 3L, 6L, 3L, 5L, 5L, 5L, 9L, 9L, 7L, 5L, 6L, 3L, 8L, 4L, 3L, 8L, 0L, 4L, 0L, 2L, 2L, 3L, 7L, 6L, 4L, 2L, 4L, 1L, 6L, 2L, 1L, 5L, 8L, 1L, 8L, 0L, 0L, 0L]

Alternative solution, also 61 bytes:

lambda n:[str(5**int(n/.7-~i))[~n]for i in range(2**n/2)]or 5

Explanation:

def f(n):
    if n == 0:
        return 5
    r = 2**n / 2
    d = 10**n 
    m = int(n/.7 + 1)
    for i in range(r):
        yield (5**(m+i)/d) % 10

The range(2**n/2) uses the observation that each cycle has length R = 2^n-1, so we just compute the n-th digits for 5^m to 5^{m + r - 1}.

The start of the cycle 5^m is the first number larger than 10ⁿ. Solving 5^m ≥ 10ⁿ gives m ≥ n / log₁₀ 5. Here we approximate log₁₀ 5 ≈ 0.7 which will break down when n = 72. We could add more digits to increase the accuracy:

| approximation             | valid until        | penalty   |
|---------------------------|--------------------|-----------|
| .7                        | n = 72             | +0 bytes  |
| .699                      | n = 137            | +2 bytes  |
| .69897                    | n = 9297           | +4 bytes  |
| .698970004                | n = 29384          | +8 bytes  |
| .6989700043               | n = 128326         | +9 bytes  |
| .6989700043360189         | too large to check | +15 bytes |
| import math;math.log10(5) | same as above      | +23 bytes |

The / d % 10 in the loop simply extract the desired digit. The alternative solution uses string manipulation. I used the trick ~n == -n-1 here to remove 1 byte.

When n = 0, the loop will give an empty list, so we special-case it by returning the hard-coded "5" when the list is empty.