$\newcommand\ep\varepsilon\newcommand\de\delta$ 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 \\ & <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*}\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).