$\newcommand{\ep}{\epsilon}$Somehow, I have only now recalled about Latala's inequalities for moments of the sums of positive independent random variables (r.v.'s), which, in particular, allow one to easily obtain the order of magnitude of $E|S_n|$.
Indeed, by the Marcinkiewicz--Zygmund inequalities,
\begin{equation*}
E|S_n|\asymp E\Big(\sum_1^n X_i^2\Big)^{1/2},
\end{equation*}
where $a\asymp b$ means that $a/C\le b\le Ca$ for some universal real constant $C>0$.
By Latala's Theorem 1 (with $X_i^2$ in place of $X_i$ in that theorem),
\begin{equation*}
E\Big(\sum_1^n X_i^2\Big)^{1/2}\asymp b_n^{1/2},
\end{equation*}
where
\begin{equation*}
b_n:=\inf\{t>0\colon E\sqrt{1+X^2/t}\le e^{1/(2n)}\} \tag{-1}
\end{equation*}
and $X:=X_1$.
Thus,
\begin{equation*}
E|S_n|\asymp b_n^{1/2}, \tag{0}
\end{equation*}
which determines $E|S_n|$ up to a universal real constant factor.
In particular, this implies
\begin{equation*}
E\Big|\frac{S_n}n\Big|\to0\tag{1}
\end{equation*}
whenever $E|X|<\infty$.
Indeed, for real $t>0$, let
\begin{equation*}
Y_t:=(\sqrt{1+X^2/t}-1)\sqrt t.
\end{equation*}
Then $0\le Y_t\le|X|$ and $Y_t\to0$ (since $0\le Y_t\le X^2/(2\sqrt t))$. So, by the condition $E|X|<\infty$ and dominated convergence,
\begin{equation*}
EY_t\to0
\end{equation*}
as $t\to\infty$. So, for each real $\ep>0$ and all large enough $n$ (depending on $\ep$), letting $t=(2\ep n)^2$, we have $EY_t\le\ep$, that is,
\begin{equation*}
E\sqrt{1+X^2/t}\le1+\ep/\sqrt t=1+1/(2n)<e^{1/(2n)}.
\end{equation*}
In view of (-1), this yields
\begin{equation*}
b_n\le(2\ep n)^2.
\end{equation*}
So, (0) implies
\begin{equation*}
\limsup_n E\Big|\frac{S_n}n\Big|\le2C\ep
\end{equation*}
for some real $C>0$ and every $\ep>0$. Thus, (1) follows.
Using (0), one can show that actually the bound $c_n$ on $E\big|\frac{S_n}n\big|$ in formula (2) in my other answer here is optimal up to a universal real constant factor, that is, one has
\begin{equation*}
E|S_n|\asymp nc_n, \tag{2}
\end{equation*}
provided that $P(X=0)<1$.
Indeed, by Jensen's inequality, $E|S_n|=E|S_n^X|\le E|S_n^Z|=E|S_n^X-S_n^Y|\le2E|S_n|$, where $S_n^X,S_n^Y,S_n^Z$ are as in my other answer on this page. So, (2) can be rewritten as
\begin{equation*}
E|S_n^Z|\asymp nc_n. \tag{2Z}
\end{equation*}
Also, by (0),
\begin{equation*}
E|S_n^Z|\asymp\sqrt{b_n^Z}, \tag{0Z}
\end{equation*}
where $b_n^Z$ is obtained from $b_n$ by replacing $X$ by $Z:=Z_1$ in the definition (-1) of $b_n$.
Next, for real $u$
\begin{equation*}
\sqrt{1+u^2}-1\asymp\min(u^2,|u|)=u^2\,1(|u|\le1)+|u|\,1(|u|>1).
\end{equation*}
On the other hand (cf. (-1)),
\begin{equation*}
E\sqrt{1+Z^2/b_n^Z}-1=e^{1/(2n)}-1\asymp1/n,
\end{equation*}
so that
\begin{equation*}
E\frac{Z^2}{b_n^Z}\,1(|Z|\le b_n^Z)+E\frac{|Z|}{\sqrt{b_n^Z}}\,1(|Z|>b_n^Z)\asymp\frac1n,
\end{equation*}
whence
\begin{equation*}
EZ^2\,1(|Z|\le b_n^Z)\ll\frac{{b_n^Z}}n,\quad
E|Z|\,1(|Z|>b_n^Z)\ll\frac{\sqrt{b_n^Z}}n,
\end{equation*}
where $a\ll b$ means $a\le Cb$ for some universal positive real constant $C$.
So, by formula (2) in my other answer on this page,
\begin{align*}
nc_n&\le K(b_n^Z)\sqrt n+nL(b_n^Z) \\
&=\sqrt{EZ^2\,1(|Z|\le b_n^Z)}\sqrt n+n E|Z|\,1(|Z|>b_n^Z) \\
&\ll\sqrt{\frac{{b_n^Z}}n}\sqrt n+n \frac{\sqrt{b_n^Z}}n
=2\sqrt{b_n^Z}.
\end{align*}
Therefore and in view of formulas (0Z) above and (2) in my other answer on this page,
\begin{equation*}
nc_n\ll\sqrt{b_n^Z}\asymp E|S_n^Z|\le2E|S_n|\ll nc_n.
\end{equation*}
Thus, formula (2) in this answer has been proved.
