Compactness number for a fragment of second-order logic

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 4 (previous edits done over): Now regarding lower bounds (and cf. Noah's comments below)...

Say that a (possibly beyond 1st order) logic (I'm not sure of what the formal definition of "logic" is here, but it doesn't really matter; I only use it in a limited sense below which is clear enough) is $\kappa$-compact iff whenever $T$ is a set-sized theory in that logic and all subtheories of size ${<\kappa}$ have a model, then $T$ also has a model.

Now consider the logic $\mathcal{L}$ in the (first order) language of set theory, where we have available a proper class of constant symbols, and an extra predicate symbol (which can be taken to code multiple predicates/functions below), together with the second-order statement ``I am wellfounded''. (Remark: I mistakenly said "one" constant symbol in an earlier version; but I used set-many. Actually for the purposes below we can restrict the number of constants available to something like the cardinality of $V_{\kappa+2}$, where $\kappa$ is the cardinal in question.)

Claim: Let $\kappa$ be a cardinal and suppose that $\mathcal{L}$ as above is $\kappa$-compact. Then there is a measurable cardinal $\leq\kappa$.

Proof: First suppose that $\kappa$ is inaccessible. Consider the $\mathcal{L}$-theory $T$ which includes the full first order theory of $V_{\kappa+1}$ in parameters, the formulas “$\alpha < \dot{\mu} < \kappa$”, for each $\alpha < \kappa$, where $\dot{\mu}$ is a new constant, and the (2nd order) formula “I am wellfounded”. If $M$ is a (wellfounded) model of $T$, note that there is an elementary $j:V_{\kappa+1}\to M$, the critical point $\mathrm{crit}(j)$ of $j$ exists, and is a measurable cardinal $\leq\kappa$. So we just need to see the small subtheories have models, but this follows easily from the inaccessibility of $\kappa$ (consider elementary substructures of $V_{\kappa+1}$).

Now suppose $\kappa$ is singular. Let $f:\eta\to\kappa$ be cofinal, where $\eta<\kappa$. Consider the theory $T$ consisting of the first order theory in parameters of $V_\kappa$, and using function symbols $\dot{f},\dot{g}$, the statement "$\dot{f}:\eta\to\mathrm{Ord}$ is cofinal", the equations "$\dot{f}(\alpha)=\beta$" for each $\alpha,\beta$ such that $f(\alpha)=\beta$, and the statement "$\dot{g}$ is a surjection from some ordinal onto $\mathrm{Ord}$", and (2nd order) "I am wellfounded". Then for each $\theta<\kappa$, each fragment $\bar{T}$ of size $\theta$ is satisfiable, since we may assume $\eta<\theta$, and we can find an elementary substructure of $(X,f\cap X)\preceq(V_\kappa,f)$ of cardinality $\theta$, with $\theta+1\subseteq X$, and then the transitive collapse $(M,\bar{f},g)$ is a model, where we take $g:\theta\to\mathrm{Ord}^M$ a surjection. So we get a wellfounded model $(M,f',g')$ of the full theory. Let $j:V_\kappa\to M$ be the resulting elementary embedding. Note that $\mathrm{crit}(j)<\kappa$, because otherwise $f'=f$, so by cofinality, $\kappa=\mathrm{Ord}^M$, so $M=V_\kappa$, so $g':\gamma\to\kappa$ is a surjection for some $\gamma<\kappa$, contradiction. But then $\mathrm{crit}(j)<\kappa$ is measurable.

Finally suppose that $\gamma<\kappa\leq 2^\gamma$. Then consider the theory consisting of the first order theory in parameters of $\mathcal{H}_{(2^\gamma)^+}$, "I am wellfounded", and the statement "$\dot{f}:\mathcal{P}(\gamma)\to\mathrm{Ord}$ is surjective". Any sub-theory of size $<\kappa$ is satisfiable (in fact, any sub-theory of size $\leq 2^\gamma$ is satisfiable), so we have a model, and note it gives a measurable $\leq\gamma$.

(This leaves $\omega$, but it's easy to see this is impossible; also cf. Noah's comments in the question and below.)