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 together with a constant symbol, 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.

edit 2. Suppose we have compactness at $\kappa=\gamma^+$ for the language as above, with constant symbol X. Then there is a measurable cardinal $\leq\gamma$. For consider the theory consisting of the first order theory in parameters of $H=\mathcal{H}_{\kappa}$, plus “I am wellfounded”, plus “X is a wellorder of $\gamma$“ plus the statements “$\alpha<$ the ordertype of X”, for each $\alpha<\kappa$. Clearly each size $<\kappa$ subtheory is satisfiable. But if $M$ satisfies the whole theory then $M\neq H$, because of $X^M$, and we have an elementary $\pi:H\to M$, and it has a critical point, which is measurable.