I think that if you have a logic $\mathcal{L}$ which has downward Lowenheim-Skolem (for theories of arbitrary cardinality, i.e. if $T$ has cardinality $\lambda$ and $N$ is a model $A$ of size $\theta\geq\lambda$, and $\lambda\leq\gamma\leq\theta$, then we can find a sub-model of size $\gamma$ which is elementary in $A$ (w.r.t. the relevant formulas of the logic)), then its strong compactness number is $\leq$ the least supercompact $\kappa$.
For suppose $T$ is a theory in $\mathcal{L}$ such that all subsets of $T$ of size ${<\kappa}$ have a model. Let $\lambda$ be the cardinality of $T$; we consider $T\subseteq\lambda$. Let $j:V\to M$ with $\mathrm{crit}(j)=\kappa$ and $\mathcal{P}(\lambda)\subseteq M$ and $j\upharpoonright\lambda\in M$. Then $M$ thinks that every sub-theory of $j(T)$ of size $<j(\kappa)$ has a model. But we have $T\in M$ and $T\subseteq\lambda$, and note that $T$ is equivalent to $j``T\in M$, and this has size $\lambda<j(\kappa)$ in $M$. So $T$ has a model $B$ in $M$. But by Lowenheim-Skolem, then it has a model of size $\lambda$ in $M$. Since $\mathcal{P}(\lambda)\subseteq M$, this is truly a model in $V$.
edit. If kappa is inaccessible and we have compactness at kappa with respect to the logic which includes the full first order language of set theory plus the (second order) statement “I am wellfounded”, then there is a measurable less than or equal to kappa. For consider the theory T in the language of set theory plus the constant symbol mu-dot, which includes the full first order theory of $V_{\kappa+1}$ in parameters, the formulas “alpha < mu-dot < kappa”, for each alpha < kappa, and the (2nd order) formula “I am wellfounded”. If that has a model, then it’s truly wellfounded, and it follows there’s a measurable less or equal to kappa. So we just need to see the small subtheories have models, but this easily follows from inaccessibility.