Polynomial function defined recursively by a resultant - is it well defined? Preliminaries
Let $ n $ be an integer such that $ n \geq3 $. Denote $ \left[ n \right] \equiv \{1,2, \ldots ,n \} $. Let $ P $ be a non-empty subset of $ \left[ n \right] $ such that $ \left|P \right| \geq 3 $. Denote $ X_{P} \equiv \left( x_{i} \colon i \in P \right) $ as an ordered alphabet. Let $ \mathbb{F} $ be a field such that $ \operatorname{char} \left( \mathbb{F} \right) \neq 2$. Denote $ \mathbb{F} \left[ X_{P} \right] $ as the polynomial ring over $ \mathbb{F} $ in the (commuting) letters of $ X_{P} $.
Polynomials as functions
Let $ P_{1},P_{2} $ be non-empty subsets of $ P $ such that $ \left|P_{1} \right| = \left|P_{2} \right|$, then the polynomial $ f $ in the letters of $ X_{P_{1}} $, denoted by $ f \left( X_{P_{1}} \right) \in \mathbb{F} \left[ X_{P_{1}} \right]$, and the polynomial $ f $ in the letters of $ X_{P_{2}} $, denoted by $ f \left( X_{P_{2}} \right) \in \mathbb{F} \left[ X_{P_{2}} \right]$, will be considered equivalent as functions.
Multivariate Polynomial to univariate polynomial and back
A (multivariate) polynomial in $ \mathbb{F} \left[ X_{P} \right] $ can be uniquely written as a (univariate) polynomial in the letter/variable $ x_j $ for some $ j \in P $, or in other words as a polynomial in $ \left( \mathbb{F} \left[ X_{P \setminus \{ j \}} \right] \right) \left[ x_{j} \right] $. Conversely, a polynomial in  $ \left( \mathbb{F} \left[ X_{P \setminus \{ j \}} \right] \right) \left[ x_{j} \right] $ can be uniquely written as a polynomial in $ \mathbb{F} \left[ X_{P} \right] $. Denote $ f \left( X_{P \setminus \{ j \}} ;x_{j} \right) $ as the polynomial in $ \left( \mathbb{F} \left[ X_{P \setminus \{ j \}} \right] \right) \left[ x_{j} \right] $ which is equivalent to the polynomial $ f \left( X_{P} \right) $, for any $ f \left( X_{P} \right) \in \mathbb{F} \left[ X_{P} \right]$.
Univariate resultants and multivariate polynomials
For any $ j \in P $ and $ k \in P \setminus \{ j \} $, consider the polynomials $ f \left( X_{P} \right) \in \mathbb{F} \left[ X_{P} \right]$ and $ g \left( X_{\left( \left[ n \right] \setminus P \right) \cup \{ j \} \cup \{ k \}} \right) \in \mathbb{F} \left[ X_{\left( \left[ n \right] \setminus P \right) \cup \{ j \} \cup \{ k \} }\right]$ and denote $ y \equiv x_{j} \equiv x_{k} $. Then the (univariate) resultant of the polynomials $ f \left( X_{P \setminus \{ j \}} ;y \right) \in \left( \mathbb{F} \left[ X_{P \setminus \{ j \}} \right] \right) \left[ y \right]$ and $ g \left( X_{\left( \left[ n \right] \setminus P \right) \cup \{ j \}} ; y \right) \in \left( \mathbb{F} \left[ X_{\left( \left[ n \right] \setminus P \right) \cup \{ j \} }\right] \right) \left[ y \right]$, with respect to the variable $ y $, is well-defined, and shall be denoted by
$$ \operatorname{res}_{y} \left( f \left( X_{P \setminus \{ j \}} ;y \right) ,g \left( X_{\left( \left[ n \right] \setminus P \right) \cup \{ j \}} ; y \right) \right) $$
Recursive definition
The polynomial function $ q \left( X_{ \left[ n \right]} \right) \in \mathbb{F} \left[ X_{\left[ n \right]} \right] $ is defined recursively by the condition $$ q \left( X_{ \left[ 3 \right]} \right) \equiv -x_{1}^2-x_{2}^2-x_{3}^2+2x_{1}x_{2}+2x_{1}x_{3}+2x_{2}x_{3} $$ and the rule $$ q \left( X_{ \left[ n \right]} \right) = \operatorname{res}_{y} \left( q \left( X_{P \setminus \{ j \}} ;y \right) ,q \left( X_{\left( \left[ n \right] \setminus P \right) \cup \{ j \}} ; y \right) \right)$$ for $ n \geq 4 $, any non-empty $ P \subseteq \left[ n \right] $ such that $ \left|P \right| \geq 3 $ and any $ j \in P $.
Example
Suppose that $ n = 4 $, $ P = \{1,2,3\} $ and $ j = 3 $. Then $ \left[ n \right] = \{1,2,3,4\} $, $ \left| P \right| = 3 \geq 3 $ and $ j \in P $; furthermore
$$
\begin{align}
q \left( X_{P \setminus \{ j \}} ;y \right)  & = q \left( X_{\{1,2,3\} \setminus \{ 3 \}} ;y \right)  \\
 & = q \left( X_{\{1,2\}} ;y \right) \\ 
 & = -y^2+2 \left(x_{1}+x_{2} \right)y- \left( x_{1}-x_{2} \right)^2 
\end{align}
$$
and
$$
\begin{align}
q \left( X_{\left( \left[ n \right] \setminus P \right) \cup \{ j \}} ; y \right)  & = q \left( X_{\left( \{1,2,3,4\} \setminus \{1,2,3\} \right) \cup \{ 3 \}} ;y \right)  \\
 & = q \left( X_{\{3,4\}} ;y \right) \\ 
 & = -y^2+2 \left(x_{3}+x_{4} \right)y- \left( x_{3}-x_{4} \right)^2 
\end{align}
$$
Therefore
$$
q \left( X_{ \left[ 4 \right] } \right) = \operatorname{res}_{y} \left( q \left( X_{ \{1,2 \}} ;y \right) ,q \left( X_{ \{3,4 \}} ; y \right) \right) 
$$
is unambiguous.
Is this well-defined?

Is the polynomial function defined above well-defined, in the sense that it is invariant (not even up to sign) of the choice of a subset $ P $ and its element $ j $?

Small examples are evidence for the positive answer.
Some ideas
It is not hard to show by induction that the degree of $ q \left( X_{ \left[ n \right] };y \right) \in \left( \mathbb{F} \left[ X_{\left[ n \right] \setminus \{ i \}} \right] \right) \left[ y \right]$ for any $ i \in \left[ n \right] $ is $ 2^{n-2} $, so to obtain that, in general,
$$
\begin{align}
\operatorname{res}_{y} \left( q \left( X_{P \setminus \{ j \}} ;y \right) ,q \left( X_{\left( \left[ n \right] \setminus P \right) \cup \{ j \}} ; y \right) \right)
& =
\operatorname{res}_{y} \left(q \left( X_{\left( \left[ n \right] \setminus P \right) \cup \{ j \}} ; y \right)
, q \left( X_{P \setminus \{ j \}} ;y \right) \right) 
\end{align}
$$
But this is obviously not enough. Small examples are evidence to $ q \left( X_{ \left[ n \right] } \right) $ being a symmetric function in $ X_{ \left[ n \right] } $, so $ q \left( X_{ \left[ n \right] } \right) $ must be a multi-symmetric function in $ X_{P \setminus \{ j \}},X_{\left( \left[ n \right] \setminus P \right) \cup \{ j \}} $, but I don't know how to prove the former claim.
Sidenotes
I have in the past posted another question regarding this object, but that discussion is irrelevant to the current question, so I will not link it here.
 A: Notice that
$$
  \let\eps\varepsilon
  q(x_1,x_2,x_3)
  =-\bigl(\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}\bigr) \bigl(-\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}\bigr) \bigl(\sqrt{x_1}-\sqrt{x_2}+\sqrt{x_3}\bigr) \bigl(-\sqrt{x_1}-\sqrt{x_2}+ \sqrt{x_3}\bigr)\\
  =-\prod_{\eps_1,\eps_2\in\{\pm1\}}\left(
     \sqrt x_3+\sum_{I<3}\eps_i\sqrt{x_i}\right).
$$
So the roots of this polynomial (as a one in $x_3$) are exactly
$$
  \left(\sum_{i <3}\eps_i\sqrt{x_i}\right)^2
$$
(each root appears in the above list twice, but it is a simple root in $q$).
Now a straightforward induction shows that
$$
   q(x_1,\dots,x_n) 
=\prod_{\eps_1,\dots,\eps_{n-1}\in\{\pm1\}}\left(
     \sqrt x_n+\sum_{I<n}\eps_i\sqrt{x_i}\right),
$$
and the (simple) roots of this polynomial (as a one in $x_n$) are exactly
$$
  \left(\sum_{i <n}\eps_i\sqrt{x_i}\right)^2.
$$
The inductive step here is just an application of the standard formula
$$
  \mathop{\mathrm{res}}(A,B)
  =a_0^kb_0^n\prod_{i,j}(\lambda_i-\mu_j),
$$
where the $\lambda_i$ and the $\mu_j$ list all the roots of
$$
  A(x)=a_0x^n+\dots \quad\text{and}\quad
  B(x)=b_0x^k+\dots,
$$
respectively. (Here we use that the degrees are powers of two!)
Remark. Surely, the polynomial defined by the above formulas, is symmetric in all variables, since its degree (in $\sqrt{x_i}$) is divisible by 4.
