I do not know about the original sum, but here is a perspective on the problem you mention as motivation.

Suppose you want to count solutions $\mathbf{x} \in \mathbb{F}_q^{n}$ to $F(\mathbf{x})=0$ that satisfy $x_i \neq x_j$ for all $i\neq j$. One can write $$\prod_{i<j}\mathbf{1}_{x_i \neq x_j} = \Delta(\mathbf{x})^{q-1}$$ where $\Delta(\mathbf{x}) := \prod_{i<j}(x_i-x_j)$. Then $$\sum_{\mathbf{x}:\, F(\mathbf{x})=0,\, x_i \neq x_j} 1 = q^{-1} \sum_{a \in \mathbb{F}_q}\sum_{\mathbf{x} \in \mathbb{F}_q^n} \Delta(\mathbf{x})^{q-1} \psi(a F(\mathbf{x}))$$ where $\psi$ is a fixed nontrivial additive character $\mathbb{F}_q \to \mathbb{C}$. For $F(\mathbf{x})=\sum_{i=1}^{n} x_i^2 -1$ this is $$ q^{-1} \sum_{a \in \mathbb{F}_q}\psi(-a)\sum_{\mathbf{x} \in \mathbb{F}_q^n} \Delta(\mathbf{x})^{q-1} \prod_{i=1}^{n}\psi(a x_i^2).$$ This resembles in many way the famous Selberg integral.

I do not know about the original sum, but here is a perspective on the problem you mention as motivation.

Suppose you want to count solutions $\mathbf{x} \in \mathbb{F}_q^{n}$ to $F(\mathbf{x})=0$ that satisfy $x_i \neq x_j$ for all $i\neq j$. One can write $$\prod_{i<j}\mathbf{1}_{x_i \neq x_j} = \Delta(\mathbf{x})^{q-1}$$ where $\Delta(\mathbf{x}) := \prod_{i<j}(x_i-x_j)$. Then $$\sum_{\mathbf{x}:\, F(\mathbf{x})=0,\, x_i \neq x_j} 1 = q^{-1} \sum_{a \in \mathbb{F}_q}\sum_{\mathbf{x} \in \mathbb{F}_q^n} \Delta(\mathbf{x})^{q-1} \psi(a F(\mathbf{x}))$$ where $\psi$ is a fixed nontrivial additive character $\mathbb{F}_q \to \mathbb{C}$. For $F(\mathbf{x})=\sum_{i=1}^{n} x_i^2 -1$ this is $$ q^{-1} \sum_{a \in \mathbb{F}_q}\psi(-a)\sum_{\mathbf{x} \in \mathbb{F}_q^n} \Delta(\mathbf{x})^{q-1} \prod_{i=1}^{n}\psi(a x_i^2).$$ This resembles in many ways the famous Selberg integral.

(Some authors write $\Delta(\mathbf{x})$ for $\prod_{i<j}(x_i-x_j)^2$; I did not use this convention as $q$ might be even.)