This is really a comment, but (a) it's too long and (b) is ought to be attached to both of the answers. Not only does the characterization of the $\kappa$-compactness property not depend on the availability of infinite quantifier strings (as discussed in the comments on Joel's answer and as explicitly stated in Ioannis's answer), it doesn't depend on quantifiers at all. Propositional logic suffices. 

More precisely, consider propositional logic with countable conjunctions and disjunctions. Suppose $\kappa$ is a cardinal and, for any set $\Gamma$ of sentences in this logic, if every subset of size $<\kappa$ is satisfiable, then so is $\Gamma$. I claim that $\kappa$ is $\omega_1$-strongly compact, i.e., every $\kappa$-complete ultrafilter on any set $I$ can be extended to a countably complete ultrafilter.

To prove this, let $I$ and a $\kappa$-complete filter $\mathcal F$ on it be given. Consider the following set $\Gamma$ of sentences in the propositional logic described above, with a propositional variable $\bar A$ for every subset $A$ of $I$.  $\Gamma$ contains:

First, the sentences $(\bar A\land\bar B)\leftrightarrow\overline{A\cap B}$ and $\neg\bar A\leftrightarrow\overline{I-A}$, for all $A,B\subseteq I$,

Second, the sentences $\bar A$ for all $A\in\mathcal F$,

Third, the sentences $\bigwedge_{n\in\omega}\overline{A_n} \leftrightarrow \overline{\bigcap_{n\in\omega}A_n}$ for all countable sequences $(A_n)$ of subsets of $I$.

Then any subset of $\Gamma$ of cardinality $<\kappa$ is satisfiable. In fact, we can satisfy all the sentences of the first and third sorts along with any $<\kappa$ sentences of the second sort as follows. The $<\kappa$ sentences of the second sort are $\bar A$ for some $<\kappa$ elements $A$ of $\mathcal F$. As $\mathcal F$ is $\kappa$-complete, these $A$'s have a nonempty intersection (in fact, their intersection is in $\mathcal F$), so let $i$ be  a point in that intersection. Then give each propositional variable $\bar X$ the truth value "true" if $i\in X$ and "false" otherwise. It is easy to check that our subset of $\Gamma$ is satisfied by this valuation.

So, by hypothesis, there is a valuation $v$ making all of $\Gamma$ true. Define $\mathcal U\subseteq\mathcal P(I)$ to be 
$$
\mathcal U=\{A\subseteq I:v(\bar A)=\text{true}\}.
$$
Then $\mathcal U$ is an ultrafilter on $I$ because $v$ satisfies the first batch of sentences in $\Gamma$; it extends $\mathcal F$ because of the second batch; and it is countably complete becuse of the third batch.