$\DeclareMathOperator\CT{CT}$ Let $\CT_t(f(t))$ denote the constant term of the Laurent polynomial of $f(t)$.

Define the two functions $F(x_1,\dots,x_n)$ and $G(y)$ by $$F:=\prod_{i=1}^nx_i^{-1}(1-x_i)^{-2}\prod_{1\leq i<j\leq n} (1-x_i-x_j)^{-1} \qquad \text{and} \qquad G:=n!\cdot y^{-n}e^{(n+1)y+y^2/2}.$$ I like to ask:

QUESTION.Is the following true? It would be great if there is a direct way to compare these two. $$\CT_{x_1}\CT_{x_2}\cdots\CT_{x_n}\left(F(x_1,\dots,x_n)\right)=\CT_y\left(G(y)\right).$$

**NOTE 1.** The sequence on the right-hand side is available at the OEIS as A301741 with an explicit evaluation.

**NOTE 2.** Incidentally, we also have (a consequence of Han's formula proved here)
$$\CT_{x_1}\CT_{x_2}\cdots\CT_{x_n}\left(F(x_1,\dots,x_n)\right)=
\sum_{\lambda\vdash n}f^{\lambda}\prod_{u\in\lambda}
\frac{(n+2)^{h_u}+n^{h_u}}{(n+2)^{h_u}-n^{h_u}};$$
where $h_u$ is the *hook-length* of cell $u$ (in the Young diagram of $\lambda$) and $f^{\lambda}$ is the *number of Standard Young Tableau of shape $\lambda$* (given by the hook-length formula).

**NOTE 3.** A cute analogue: let $f:=\prod_{i=1}^nx_i^{-1}(1-x_i)^{-1}\prod_{1\leq i<j\leq n}(1+x_i+x_j)$ and $g:=n!\cdot y^{-n}e^{ny-y^2/2}$. Then,
$$\CT_{x_1}\CT_{x_2}\cdots\CT_{x_n}\left(f(x_1,\dots,x_n)\right)
=\CT_y(g(y)).$$
**Proof.** Fedor's reasoning applies (it'd be nice to employ Richard's too)
\begin{align*}
{\rm CT}\, f&=
[x_1\ldots x_n] \prod_i (1-x_i)^{-1}\prod_{i<j}(1+x_i+x_j) \\
&=[x_1\ldots x_n]\prod_i\exp(x_i)\prod_{i<j}\exp(x_i+x_j-x_ix_j)\\
&=[x_1\ldots x_n] \exp\left(\sum x_i+\sum_{i<j}(x_i+x_j-x_ix_j)\right)\\
&=[x_1\ldots x_n]\exp\left(n\cdot S-S^2/2\right).
\end{align*}