This is to prove the conjecture \begin{equation*} x_n\sim\sqrt3\,n^{1/2} \tag{1}\label{1} \end{equation*} (as $n\to\infty$).
(For all integers $n\ge1$,) we have \begin{equation*} h_n:=x_{n+1}-x_n=\frac n{s_n}, \tag{2}\label{2} \end{equation*} where \begin{equation*} s_n:=x_1+\cdots+x_n, \end{equation*} with $s_0:=0$. Rewrite \eqref{2} as $s_{n+1}-2s_n+s_{n-1}=\dfrac n{s_n}$ and then as $s_{n+1}s_n-2s_n^2+s_{n-1}s_n=n$ and then as $s_{n+1}(s_{n+1}-x_{n+1})-2s_n^2+s_{n-1}(s_{n-1}+x_n)=n$ and then as $s_{n+1}^2-2s_n^2+s_{n-1}^2=n+s_{n+1}x_{n+1}-s_{n-1}x_n$. Note also that $s_{n+1}x_{n+1}-s_{n-1}x_n=s_n(x_{n+1}-x_n)+x_{n+1}^2+x_n^2=n+x_{n+1}^2+x_n^2$, by \eqref{2}. So, \begin{equation*} t_n:=s_{n+1}^2-2s_n^2+s_{n-1}^2=2n+x_{n+1}^2+x_n^2. \tag{3}\label{3} \end{equation*}
It follows that
\begin{equation*}
t_n\ge2n. \tag{4}\label{4}
\end{equation*}
Suppose that
\begin{equation*}
t_n\gtrsim cn \tag{5}\label{5}
\end{equation*}
for some real $c>0$. By \eqref{3}, the $t_n$'s are the second (symmetric) differences of the $s_n^2$'s. So, by \eqref{5},
\begin{equation*}
s_n^2\gtrsim\frac c6\,n^3\quad\text{and hence}\quad s_n\gtrsim\sqrt{\frac c6}\,n^{3/2}. \tag{6}\label{6}
\end{equation*}
So, by \eqref{2},
\begin{equation*}
h_n\lesssim \sqrt{\frac6c}\,n^{-1/2} \quad\text{and hence}\quad
x_n\lesssim \sqrt{\frac6c}\,2n^{1/2}. \tag{7}\label{7}
\end{equation*}
So, by \eqref{3},
\begin{equation*}
t_n\lesssim 2n+2\frac6c\,4n=\Big(2+\frac{48}c\Big)n. \tag{8}\label{8}
\end{equation*}
So (cf. \eqref{6}),
\begin{equation*}
s_n^2\lesssim\Big(2+\frac{48}c\Big)\frac{n^3}6
=\Big(\frac13+\frac8c\Big)n^3
\quad\text{and hence}\quad
s_n\lesssim\sqrt{\frac13+\frac8c}\,n^{3/2}. \tag{9}\label{9}
\end{equation*}
So, by \eqref{2},
\begin{equation*}
h_n\gtrsim\frac1{\sqrt{\frac13+\frac8c}}\,n^{-1/2} \quad\text{and hence}\quad
x_n\gtrsim \frac2{\sqrt{\frac13+\frac8c}}\,n^{1/2}. \tag{10}\label{10}
\end{equation*}
So, by \eqref{3},
\begin{equation*}
t_n\gtrsim 2n+2\frac4{\frac13+\frac8c}n=f(c)n \tag{11}\label{11}
\end{equation*}
(whenever \eqref{5} holds), where
\begin{equation*}
f(c):=2+\frac8{\frac13+\frac8c}.
\end{equation*}
It follows from \eqref{4} that for all integers $k\ge0$ \begin{equation*} t_n\gtrsim c_kn, \tag{12}\label{12} \end{equation*} where \begin{equation*} c_0:=2 \end{equation*} and \begin{equation*} c_{k+1}:=f(c_k). \end{equation*} The function $f$ is continuously increasing on $[2,\infty)$. Also, $f(c)>c$ for $c\in[2,8)$ and $f(c)<c$ for $c\in(8,\infty)$. It follows that $c_k\uparrow 8$ as $k\to\infty$. So, by \eqref{12}, \begin{equation*} t_n\gtrsim8n; \end{equation*} that is, \eqref{5} holds with $c=8$. So, by \eqref{7} and \eqref{10}, \begin{equation*} x_n\lesssim \sqrt{\frac68}\,2n^{1/2}=\sqrt3\,n^{1/2} \quad\text{and}\quad x_n\gtrsim \frac2{\sqrt{\frac13+\frac88}}\,n^{1/2}=\sqrt3\,n^{1/2}. \end{equation*} Thus, \eqref{1} is proved. $\quad\Box$
It certainly seems that this method will work as well in the more general case when the numerator $n$ in \eqref{2} is replaced by $n^p$ for real $p>0$. However, I have not checked the details.