Let $n \in \mathbb{N}_{\geq 2}.$ For any real numbers $a_1,a_2,...,a_n$ denote $S_0=1$ and for $1 \leq k \leq n$ denote
$$S_k=\sum_{1 \leq i_1 < i_2 < ... <i_k \leq n}a_{i_1}a_{i_2}...a_{i_k}$$
Find the number of $n-$tuples $(a_1,a_2,...a_n)$ such that $$(S_n-S_{n-2}+S_{n-4}-...)^2+(S_{n-1}-S_{n-3}+S_{n-5}-...)^2=2^nS_n.$$
1. We start by analyzing the given equation:
   \[
   (S_n - S_{n-2} + S_{n-4} - \ldots)^2 + (S_{n-1} - S_{n-3} + S_{n-5} - \ldots)^2 = 2^n S_n
   \]
   Here, \( S_k \) represents the sum of all products of \( k \) distinct elements from the set \( \{a_1, a_2, \ldots, a_n\} \).

2. Consider the polynomial \( P(x) = (x - a_1)(x - a_2) \cdots (x - a_n) \). The coefficients of this polynomial are related to the \( S_k \) terms:
   \[
   P(x) = x^n - S_1 x^{n-1} + S_2 x^{n-2} - \cdots + (-1)^n S_n
   \]
   where \( S_k \) is the sum of the products of \( k \) distinct \( a_i \)'s.

3. The given equation can be interpreted in terms of the roots of the polynomial \( P(x) \). Specifically, we need to find the \( n \)-tuples \( (a_1, a_2, \ldots, a_n) \) such that the equation holds.

4. Let's consider the case where all \( a_i \)'s are either 1 or -1. This simplifies the problem because the sums \( S_k \) will be easier to compute.

5. If \( a_i = 1 \) or \( a_i = -1 \), then:
   \[
   a_i^2 = 1 \quad \text{for all } i
   \]
   This implies that:
   \[
   S_n = \prod_{i=1}^n a_i
   \]
   and
   \[
   S_{n-2} = \sum_{1 \leq i < j \leq n} \prod_{k \neq i,j} a_k
   \]
   and so on.

6. The given equation simplifies to:
   \[
   (S_n - S_{n-2} + S_{n-4} - \ldots)^2 + (S_{n-1} - S_{n-3} + S_{n-5} - \ldots)^2 = 2^n S_n
   \]

7. For \( a_i = 1 \) or \( a_i = -1 \), we need to check if the equation holds. Notice that:
   \[
   S_n = \prod_{i=1}^n a_i = (-1)^k \quad \text{where } k \text{ is the number of } -1 \text{'s}
   \]

8. The equation \( a_j^2 + 1 = 2|a_j| \) implies \( a_j = 1 \) or \( a_j = -1 \).

9. For the equation to hold, there must be an even number of -1's. This is because the sum of squares of the terms must equal \( 2^n \), which is only possible if the number of -1's is even.

10. The first \( n-1 \) numbers can be chosen randomly to be -1 or 1, and the last one is uniquely determined by them to ensure an even number of -1's.

11. Therefore, there are \( 2^{n-1} \) solutions, as each of the first \( n-1 \) numbers can be either -1 or 1, and the last one is determined.

The final answer is \( \boxed{ 2^{n-1} } \)