Intuitively, it should approach zero fast as soon as $|\mathcal S_{\mathsf{big}}|$ is at all smaller than $n$ (even like $n/2$) because virtually all subsets $\mathcal S_{\mathsf{small}}$ will contain some element not in $\mathcal S_{\mathsf{big}}$.
(Quick edit: I am assuming that $\mathcal S_{\mathsf{big}}$ and $\mathcal S_{\mathsf{small}}$ are each drawn indpendently and uniformly at random from the set of all subsets of $\mathcal S$ having the specified size.)
Let $a = |\mathcal S_{\mathsf{big}}|$ and $b = |\mathcal S_{\mathsf{small}}|$. For any realization of $\mathcal S_{\mathsf{big}}$, there are ${a \choose b}$ possible realizations of $\mathcal S_{\mathsf{small}}$ contained in $\mathcal S_{\mathsf{big}}$. There are ${n \choose b}$ total possible realizations of $\mathcal S_{\mathsf{small}}$.
Hence the probability of containment is exactly
\begin{align}
\frac{{a \choose b}}{{n \choose b}}
&= \frac{a!}{(a-b)!} \frac{(n-b)!}{n!} \\
&= \frac{a(a-1)\cdots(a-b+1)}{n(n-1)\cdots(n-b+1)}.
\end{align}
For $b$ small compared to $n$, this is essentially $\left(\frac{a}{n}\right)^b$. That is exactly the answer you'd get if you had fixed $\mathcal S_{\mathsf{big}}$ and picked $\mathcal S_{\mathsf{small}}$ by drawing $b$ elements independently (although this method would create duplicates).
More precisely, $\left(\frac{a}{n}\right)^b$ is an upper bound on $\Pr[\mathcal S_{\mathsf{small}} \subseteq \mathcal S_{\mathsf{big}}]$, and a lower bound is $\left(\frac{a-b}{n-b}\right)^b$.
For your examples, we get probabilities upper-bounded by $\left(\frac{n^{1-s}}{n}\right)^{n^s} = n^{-s n^s}$ and by $(\log n)^{-s(\log n)^s}$. Of course both of these approach zero fast.
A necessary condition for a limit of $\delta > 0$ as $n,a,b \to \infty$ is that $\left(\frac{a}{n}\right)^b \geq \delta$. I think this implies something like $a = n\left(1 - \frac{O(1)}{b}\right)$.
