$\newcommand\ep\varepsilon\newcommand\de\delta$
Let us show more: for all $x\in A$,
\begin{equation*}
\frac{F_{\ep}((n))}{F_{\ep}(x)}\to0\tag{$*$}
\end{equation*}
(as $\ep\downarrow0$).

Indeed, take any $x\in A$ and let
\begin{equation*}
y_n:=x_n/n^2,
\end{equation*}
so that $\sum_n y_n<\infty$ and
\begin{equation*}
F_{\ep}(x)=\sum_{n=1}^\infty 2^{-\ep n^2 y_n}.
\end{equation*}
By Jensen's inequality for the convex function $u\mapsto2^{-u}$, for any natural $N$
\begin{equation*}
F_{\ep}(x)\ge\sum_{n=1}^N 2^{-\ep n^2 y_n}\ge N2^{-\ep \sum_1^N n^2 y_n/N}. \tag{1}
\end{equation*}
Take now any real $\de>0$. Then, by the condition $\sum_n y_n<\infty$, there is a natural $M_\de$ such that $\sum_{n>M_\de} y_n<\de/2$. So, for $N>M_\de$,
\begin{equation*}
\sum_1^N n^2 y_n=\sum_{n\le M_\de} n^2 y_n+\sum_{M_\de<n\le N} n^2 y_n
\le\sum_{n\le M_\de} n^2 y_n+N^2\de/2<N^2\de
\end{equation*}
if we also have $N^2>2\sum_{n\le M_\de} n^2 y_n/\de$, and then, by (1),

\begin{equation*}
F_{\ep}(x)\ge N2^{-\ep N\de}.
\end{equation*}
Choosing now $N\sim\dfrac1{\ep\de}$ with $\ep\downarrow0$, we have
\begin{equation*}
F_{\ep}(x)\ge\dfrac1{3\ep\de},
\end{equation*}
for each real $\de>0$ and all small enough $\ep>0$.

On the other hand,
\begin{equation*}
F_{\ep}((n))=\sum_{n=1}^\infty 2^{-\ep n}=\frac{2^{-\ep}}{1-2^{-\ep}}\sim\frac1{\ep\ln2}.
\end{equation*}

Now ($*$) follows.