Partial progress: here is a proof for the analogous "Gaussian case" (i.e., $Z\sim N(\mathbf{0}_d, \mathbf{I}_d)$) and $n=2$. I was hoping to get some inspiration for it to handle the "Rademacher" $Z$ case I am interested in, though it doesn't seem to be helpful after all...
The goal is to bound an analogous quantity as (1), which for $n=2$ is
$$
\psi(\gamma,d,2) = \frac{e^{-\gamma^2 d}}{(1-2\gamma^2)^{d/2}}-2 + e^{\gamma^2d}\mathbb{E}_{XY}\left[ {\frac{\mathbb{E}_{Z}[ (1+\gamma\langle X,{Z}\rangle)(1+\gamma\langle Y,{Z}\rangle) ]^2}{ e^{\frac{\gamma^2}{2}\lVert X+Y \rVert_2^2} } }\right]
$$
(the denominator changes because it comes from $\mathbb{E}_Z[e^{\gamma \langle\sum_{j=1}^n X^{(j)},Z\rangle}]$, and that expression changes between the Gaussian and Rademacher cases. Similarly for why the additive "-1" is now a less nice expression.)
We can expand and compute the numerator (before squaring) as
\begin{align*}
\mathbb{E}_{Z}\left[(1+\gamma\langle X, Z\rangle)(1+\gamma\langle Y, Z\rangle) \right]
= \mathbb{E}_{Z}\left[1+\gamma\langle{X+Y,Z}\rangle + \gamma^2\langle X, Z\rangle\langle Y, Z\rangle \right]
= 1+ \gamma^2\langle X, Y\rangle
\end{align*}
since $Z$ is Gaussian. From there, observing that $V := \frac{\lVert X+Y\rVert_2^2}{2}$ is a $\chi^2(d)$ r.v., we get
\begin{align*}
e^{-\gamma^2d }&(1+\psi(d ,\gamma,2))\\
&= \mathbb{E}_{XY}[{ e^{-\frac{\gamma^2}{2}\lVert X+Y\rVert_2^2} (1+ \gamma^2\langle X, Y\rangle)^2 }]\\
&= \mathbb{E}_{V}[{ e^{-\gamma^2 V} }] + 2\gamma^2 \mathbb{E}_{XY}[{\langle X, Y\rangle e^{-\frac{\gamma^2}{2}\lVert X+Y\rVert_2^2} }] + \gamma^4\mathbb{E}_{XY}[{\langle X, Y\rangle^2e^{-\frac{\gamma^2}{2}\lVert X+Y\rVert_2^2} }] \\
&\leq \mathbb{E}_{V}[{ e^{-\gamma^2 V} }] + 2\gamma^2 \mathbb{E}_{XY}[{\langle X, Y\rangle e^{-\frac{\gamma^2}{2}\lVert X+Y\rVert_2^2} }] + \gamma^4\mathbb{E}_{XY}[{\langle X, Y\rangle^2}] \\
&= \frac{1}{(1+2\gamma^2)^{d /2}} + \gamma^2 \mathbb{E}_{XY}[{(\lVert X+Y\rVert_2^2 - \lVert X\rVert_2^2 - \lVert Y\rVert_2^2) e^{-\frac{\gamma^2}{2}\lVert X+Y\rVert_2^2} }] + d \gamma^4\,.
\end{align*}
The second term can be computed explicitly. Indeed, we have
\begin{align*}
\mathbb{E}_{XY}[{\lVert X+Y\rVert_2^2e^{-\frac{\gamma^2}{2}\lVert X+Y\rVert_2^2} }]
&= 2 \mathbb{E}_{V}[{V e^{-\gamma^2V} }] = \frac{2d }{(1+2\gamma^2)^{1+d /2}} \\
\mathbb{E}_{XY}[{\lVert X\rVert_2^2e^{-\frac{\gamma^2}{2}\lVert X+Y\rVert_2^2} }]
&= \mathbb{E}_{X}[{\lVert X\rVert_2^2 \mathbb{E}_{Y}[{ e^{-\frac{\gamma^2}{2}\lVert X+Y\rVert_2^2} }] }]
= \mathbb{E}_{X}[{\lVert X\rVert_2^2 \prod_{i=1}^d \mathbb{E}_{Y}{ e^{-\frac{\gamma^2}{2}(X_i+Y_i)^2} } }]\\
&= {(1+\gamma^2)^{-d/2}}\mathbb{E}_{X}[{\lVert X\rVert_2^2e^{-\frac{\gamma^2}{2(1+\gamma^2)}\lVert X\rVert_2^2} }]
= \frac{d (1+\gamma^2)}{(1+2\gamma^2)^{1+d /2}}
\end{align*}
from which the second term equals
\begin{align*}
\mathbb{E}_{XY}[{(\lVert X+Y\rVert_2^2 - \lVert X\rVert_2^2 - \lVert Y\rVert_2^2) e^{-\frac{\gamma^2}{2}\lVert X+Y\rVert_2^2} }]
&= - \frac{2d \gamma^2}{(1+2\gamma^2)^{1+d /2}}\,.
\end{align*}
Overall, we get
\begin{align*}
e^{-\gamma^2d }\left(2-\frac{e^{-\gamma^2 d}}{(1-2\gamma^2)^{d/2}}+\psi(d ,\gamma,2)\right)
&\leq \frac{1}{(1+2\gamma^2)^{d /2}}\left(1- \frac{2d \gamma^4}{1+2\gamma^2}\right) + d \gamma^4
\leq \frac{1}{(1+2\gamma^2)^{d /2}} + d \gamma^4\,,
\end{align*}
from which
\begin{align*}
\psi(d ,\gamma,2)
= O( d \gamma^4 )
\end{align*}
as long as $\gamma^2d = O(1)$.
