Great question! This is not a complete answer, but hopefully it gets the ball rolling . . .
Theorem: $\mathfrak s_{-}\geq \mathrm{add}(\mathcal M)$, where $\mathrm{add}(\mathcal M)$ is the smallest size of a collection of meager sets whose union is non-meager.
Proof: Let $X$ be a set of reals with $|X| < \mathrm{add}(\mathcal M)$, and let $\varepsilon_n$, $n < \omega$, be a sequence of numbers. We want to show how to cover $X$ with intervals $I_n$ of length $\varepsilon_n$.
The basic idea is to use the following fact about Cohen forcing: in a Cohen extension, the set of ground model reals has strong measure zero. This was proved by Martin Goldstern in his first-ever MathOverflow post. Here are the details for the rest of the argument.
Let $M$ be a countable transitive model of set theory with $X \in M$, $X \subseteq M$, and the sequence of the $\varepsilon_n$ in $M$. (Here a "model of set theory" means the usual thing -- only finitely many of the ZFC axioms, but enough to make my argument work!) Such a model $M$ exists by the Löwenheim-Skolem-Tarski Theorem.
Because $|M| < \mathrm{add}(\mathcal M)$, there is a real $c$ that is Cohen over $M$. This uses the fact that a real is Cohen over $M$ if and only if it is not in any meager Borel set coded inside of $M$. Because fewer than $\mathrm{add}(\mathcal M)$ meager Borel sets are coded in $M$, there is (by the definition of $\mathrm{add}(\mathcal M)$) some real missing them all.
By Martin's argument that I linked to earlier, in $M[c]$ we have "the set $X$ is strong measure zero". But I specifically put the sequence $\varepsilon_n$ into $M$, which means that this sequence is also in $M[c]$. Thus, in $M[c]$, we have "there is a sequence $I_n$ of intervals of lengths $\varepsilon_n$ covering $X$". But this statement is upward absolute: if it's true in $M[c]$, then it's true. QED