$\newcommand\ep\varepsilon\newcommand\de\delta$
Let us show more:
\begin{equation*}
\frac{F_{\ep}(x)}{F_{\ep}((1/n))}\to0\tag{$*$}
\end{equation*}
(as $\ep\downarrow0$).
Indeed,
\begin{equation*}
F_{\ep}((1/n))=\sum_{i=1}^\infty 2^{-\ep i}=\frac{2^{-\ep}}{1-2^{-\ep}}\asymp\frac1\ep. \tag{0}
\end{equation*}
On the other hand, take any positive $x_n$'s such that $\sum_1^\infty x_n<\infty$. For each natural $k$, let $J_k$ denote the set of all natural $n$ such that $\frac1k\le x_n<\frac1{k-1}$, where $\frac1{k-1}:=\infty$ for $k=1$:
\begin{equation*}
n\in J_k\iff\frac1k\le x_n<\frac1{k-1}.
\end{equation*}
Then the $J_k$'s partition the set of all natural numbers. Moreover, the condition $\sum_1^\infty x_n<\infty$ implies
\begin{equation*}
\sum_{k=1}^\infty|J_k|/k<\infty, \tag{1}
\end{equation*}
where $|J_k|$ is the cardinality of $J_k$.
In particular, it follows that $|J_k|<\infty$ for all $k$.
Further, $2^{-\ep/x_n}<2^\ep\times2^{-\ep k}$ for $n\in J_k$. So,
\begin{equation*}
F_{\ep}(x)<2^\ep\sum_{k=1}^\infty 2^{-\ep k}|J_k|.
\end{equation*}
Take now any real $\de>0$ and, in view of (1), let $k_\de$ be a natural number such that
\begin{equation*}
\sum_{k\ge k_\de}|J_k|/k<\de.
\end{equation*}
Let $c_\de:=\sum_{k=1}^{k_\de-1}|J_k|$.
Then
\begin{align*}
F_{\ep}(x)&<c_\de+2^\ep\sum_{k\ge k_\de} k2^{-\ep k}|J_k|/k \\
& \le c_\de+2^\ep\max_{k\ge1}(k2^{-\ep k})\,\sum_{k\ge k_\de} |J_k|/k \\
& <c_\de+2^\ep\frac1{\ep e\ln2}\,\de \\
& <c_\de+\de/\ep
\end{align*}
if $\ep\in(0,1/2)$.
So,
\begin{equation}
\limsup_{\ep\downarrow0}\frac{F_{\ep}(x)}{1/\ep}\le\de,
\end{equation}
for every real $\de>0$. Now ($*$) follows, in view of (0).
