Determine the integers $0 \le a \le b \le c \le d$ such that: $$2^n= a^2 + b^2 + c^2 + d^2.$$
We need to determine the integers \(0 \le a \le b \le c \le d\) such that:
\[ 2^n = a^2 + b^2 + c^2 + d^2. \]

We will analyze the problem by considering the parity of \(a, b, c,\) and \(d\).

1. **Case 1: Exactly one or three of \(a, b, c, d\) are odd.**

   If exactly one or three of \(a, b, c, d\) are odd, then the sum \(a^2 + b^2 + c^2 + d^2\) is odd. This implies that \(2^n\) must be odd. However, \(2^n\) is even for all \(n \geq 1\). Therefore, this case is not possible.

2. **Case 2: Exactly two of \(a, b, c, d\) are odd.**

   If exactly two of \(a, b, c, d\) are odd, then the sum \(a^2 + b^2 + c^2 + d^2\) is even but not divisible by 4. Specifically, it is congruent to 2 modulo 4:
   \[ a^2 + b^2 + c^2 + d^2 \equiv 1 + 1 + 0 + 0 \equiv 2 \pmod{4}. \]
   This implies that \(2^n \equiv 2 \pmod{4}\), which is only possible if \(n = 1\). Therefore, we have:
   \[ 2^1 = 2 = a^2 + b^2 + c^2 + d^2. \]
   The only solution in non-negative integers is:
   \[ (a, b, c, d) = (1, 1, 0, 0) \]
   and its permutations.

3. **Case 3: All four of \(a, b, c, d\) are even.**

   If all four of \(a, b, c, d\) are even, we can write \(a = 2a_1, b = 2b_1, c = 2c_1, d = 2d_1\). Substituting these into the equation, we get:
   \[ 2^n = (2a_1)^2 + (2b_1)^2 + (2c_1)^2 + (2d_1)^2 = 4(a_1^2 + b_1^2 + c_1^2 + d_1^2). \]
   Dividing both sides by 4, we obtain:
   \[ 2^{n-2} = a_1^2 + b_1^2 + c_1^2 + d_1^2. \]
   This is a similar equation but with a smaller power of 2. We can apply the same analysis recursively.

4. **Case 4: All four of \(a, b, c, d\) are odd.**

   If all four of \(a, b, c, d\) are odd, then the sum \(a^2 + b^2 + c^2 + d^2\) is congruent to 4 modulo 8:
   \[ a^2 + b^2 + c^2 + d^2 \equiv 1 + 1 + 1 + 1 \equiv 4 \pmod{8}. \]
   This implies that \(2^n \equiv 4 \pmod{8}\), which is only possible if \(n = 2\). Therefore, we have:
   \[ 2^2 = 4 = a^2 + b^2 + c^2 + d^2. \]
   The only solution in non-negative integers is:
   \[ (a, b, c, d) = (1, 1, 1, 1) \]
   and its permutations.

To sum up, all solutions are:
\[ (2^x, 0, 0, 0), (2^x, 2^x, 0, 0), (2^x, 2^x, 2^x, 2^x) \]
and their permutations where \(x \in \mathbb{Z}^0\).

The final answer is \( \boxed{ (2^x, 0, 0, 0), (2^x, 2^x, 0, 0), (2^x, 2^x, 2^x, 2^x) } \) and their permutations where \(x \in \mathbb{Z}^0\).