This answer seems to imply that: for an ordinal $\alpha$, to be recursively inaccessible (i.e. $\alpha$ is admissible and limit of admissible) implies to be not locally countable (i.e. $L_\alpha \models \exists \beta \ ``\beta \text{ is uncountable"}$). Here is the relevant excerpt:

If there is some $r\in L_\alpha$ with $\omega_1^r=\alpha$, then $L_\alpha$ will be

locally countable(= $L_\alpha\models$ "every set is countable"). But plenty of countable admissible $\alpha$s don't give rise to locally countable levels of $L$! In particular, if $\alpha$ is an admissible limit of admissibles (= "recursively inaccessible") then every real in $L_\alpha$ is contained in some admissible $L_\beta$ with $\beta<\alpha$.

However I believe that if by "admissible" we read here "$\Sigma_1$-admissible", this statement does not hold as in this sense the first recursively inaccessible ordinal appears way before the first ordinal that is not locally countable (see resp. 2.3 and 2.21 in DA Madore's zoo of ordinals). Whence my question:

Does the statement hold if by "admissible" we read "$\Sigma_n$-admissible for all $n$"? If no, can we describe the first ordinal which is not locally countable in term of higher recursive inaccessibility?