By exchangeability, $\mathrm E\left(\frac{S_k}{S_n}\right)=\frac{k}n$ for every $0\leqslant k\leqslant n$.
Heuristics based on the functional central limit theorem suggest that, for every $0\leqslant t\leqslant 1$, when $k\approx tn$, $S_k\approx nt+\sqrt{n}\cdot B_t$ for a standard Brownian motion $(B_t)_{0\leqslant t\leqslant 1}$. Thus,
$$
\frac{S_k}{S_n}-\mathrm E\left(\frac{S_k}{S_n}\right)\approx\frac{nt+\sqrt{n}\cdot B_t}{n+\sqrt{n}\cdot B_1}-t\approx\frac1{\sqrt{n}}(B_t-tB_1).
$$
The process $(B_t-tB_1)_{0\leqslant t\leqslant 1}$ is a standard Brownian bridge.
This suggests that the expectation $\mathfrak e_n$ you are asking for behaves like $\mathrm E(M)/\sqrt{n}$, where
$$
M=\max\limits_{0\leqslant t\leqslant 1}(B_t-tB_1).
$$
Finally, since $\mathrm P(M\geqslant x)=\mathrm e^{-2x^2}$ for every $x\geqslant0$, $\mathrm E(M)=\sqrt{\pi/8}$ and all this would yield the asymptotics
$$
\lim\limits_{n\to\infty}\sqrt{n}\cdot\mathfrak e_n=\sqrt{\pi/8}=0.626657\ldots
$$
Edit The Brownian bridge is distributed like $(B_t)_{0\leqslant t\leqslant 1}$ conditioned on $B_1=0$, hence, for every $x\gt0$,
$$
\mathrm P(M\geqslant x)=\mathrm P(M_1\geqslant x\mid B_1=0),\qquad
M_1=\max\limits_{0\leqslant t\leqslant 1}B_t.
$$
Informally, intyroducing $T_x=\inf\{t\geqslant0\mid B_t\geqslant x\}$,
$[M_1\geqslant x]=[T_x\leqslant1]$, hence
$$
\mathrm P(M_1\geqslant x\mid B_1=0)\approx\left.\frac{Q(\mathrm dz)}{\mathrm P(B_1\in\mathrm dz)}\right|_{z=0}
$$
where
$$
Q(\mathrm dz)=\mathrm P(T_x\leqslant1,B_1\in\mathrm dz)=\mathrm P(T_x\leqslant1)\mathrm P(B_1\in\mathrm dz\mid T_x\leqslant1).
$$
By the reflection principle, for $z\lt x$,
$$
\mathrm P(B_1\in\mathrm dz\mid T_x\leqslant1)=\mathrm P(B_1\in\mathrm 2x-dz\mid T_x\leqslant1),
$$
hence $Q(\mathrm dz)=\mathrm P(B_1\in\mathrm 2x-dz, T_x\leqslant1)=\mathrm P(B_1\in\mathrm 2x-dz)$. Introducing the density $\varphi$ of the distribution of $B_1$ and applying this to $z=0$, this yields $\mathrm P(M\geqslant x)=\varphi(2x)/\varphi(0)$, which is the formula used above.