This answer is based on a comment by J. Kynčl. It gives a bound roughly $\rho_{min}(S) \leq \left(\frac{1}{1.074}\right)^d$ which is already exponentially decreasing, but perhaps still far from the optimum. Thus, a better bound would still be of interest.

Let $AB$ one of the longest edges of $S$. Without loss of generality, at least half of the remaining vertices of $S$ are not farther from $B$ than from $A$. Let $V_1, \dots, V_k$ be such vertices ($k \geq (d-1)/2$) and $U_1, \dots, U_{\ell}$ be the remaining vertices (closer to $A$ than to $B$). As J. Kynčl points out, the angles $V_iAB$ are at most $60^°$. We also need that angles $U_iAB$ are at most $90^°$ since $BU_i$ is at most as long as $AB$.

Let $h$ be the hyperplane perpendicular to $AB$ passing through $A$ and $h^+$ be the halfspace containing $B$ with the boundary hyperplane $h$. Let $C$ be the cone with apex $A$ determined by $S$ (that is, our task is to determine which fraction of $C$ belongs to a ball $B(A,\varepsilon)$ with center $A$ and small radius $\varepsilon$. From the discussion above it follows that $C$ is fully contained in $h^+$. Furthemore, let $\kappa$ be the affine $(k+1)$-space determined by $A$, $B$, $V_1$, $\dots$, $V_k$. We also need another $(k+1)$-dimensional cone $C_{60}$ which is formed by all points $X$ in $\kappa$ such that the angle $XAB$ is at most $60^°$. From the discussion above it follows that $V_i \in C_{60}$, and consequently $C \cap \kappa \subseteq C_{60}$. It is not too difficult to compute that $\hbox{vol}_{k+1}(B(A,\varepsilon) \cap C_{60})/\hbox{vol}_{k+1}(B(A,\varepsilon)) \leq \left(\frac{\sqrt 3}2\right)^{k+1}$, where $\hbox{vol}_{k+1}$ is the $(k+1)$-dimensional volume in $\kappa$. (See the left picture below.)

Now, let us consider a small enough ball $B(A, \varepsilon)$ and let us estimate $\hbox{vol}(B(A, \varepsilon) \cap C)/ \hbox{vol}(B(A, \varepsilon))$.
For this let us consider a $(k+1)$-space $\kappa'$ parallel with $\kappa$. The task is to show that

$$(*) \hskip{2cm} \frac{\hbox{vol}_{k+1}(B(A, \varepsilon) \cap C \cap \kappa')}{\hbox{vol}_{k+1}(B(A, \varepsilon) \cap \kappa')} \leq \left(\frac{\sqrt 3}2\right)^{k+1}.$$
As soon as we show $(*)$ we get the same bound on $\hbox{vol}(B(A, \varepsilon) \cap C)/ \hbox{vol}(B(A, \varepsilon))$ by the Fubini theorem.

In order to show $(*)$, let us first reailize that $C \cap \kappa'$ is either empty or it equals to $(C \cap \kappa) + Y$, where $Y$ is the intersection point of $\kappa'$ and the $\ell$-dimensional cone determined by $A$ and $U_1, \dots, U_\ell$ (here, for simplicity, we assume that $A$ is the origin). Thus, in particular, $Y \in h^+$ and $C \cap \kappa' \subseteq Y + C_{60}$. (See the right picture above.)

The final step is thus to show that $\hbox{vol}_{k+1}(B(A, \varepsilon) \cap (Y + C_{60})) \leq \hbox{vol}_{k+1}(B(A, \varepsilon) \cap (Z + C_{60}))$ where $Z$ is the center of $B(A, \varepsilon) \cap \kappa'$. This inequality is best shown by the picture below. (First shift $Y$ to $Y'$ on $h$. Then bound $\hbox{vol}_{k+1}(B(A, \varepsilon) \cap (Y' + C_{60}))$ by decomposing $B(A, \varepsilon) \cap (Y' + C_{60})$ into two parts as on the middle and right picture.)

Finally, $\hbox{vol}_{k+1}(B(A, \varepsilon) \cap (Z + C_{60}))/\hbox{vol}_{k+1}(B(A, \varepsilon)) \leq \left(\frac{\sqrt 3}2\right)^{k+1}$ as in the case of $C_{60}$. This gives the final bound
$$\rho_{min}(S) \leq \left(\frac{\sqrt 3}2\right)^{(d+1)/2} \leq \left(\frac{1}{1.074}\right)^d.$$