Assuming that $n > 0,$ let $\sigma_n$ denote the smallest ordinal $\alpha$ such that $L_\alpha \prec_{\Sigma_n} L_{\omega_1}.$
Let $\tau_0$ denote the supremum of the set $\{\sigma_n : n \in \omega\}$.
Let $\tau_1$ denote the smallest ordinal $\alpha$ such that $L_\alpha \prec L_{\omega_1}$ (i.e. $L_\alpha$ is fully elementary in $L_{\omega_1}$).
Question: what is the difference (if any) between these ordinals? Which of the two is larger?
$\tau_0 = \tau_1$.
Let's get the trivialities out of the way first: Since for all $\alpha$, $L_{\alpha}\prec L_{\omega_1}$ implies $L_{\alpha}\prec_{\Sigma_n}L_{\omega_1}$, we have $\sigma_n\leq \tau_1$ for all $n$. Thus $\tau_0 = \sup_{n\in \omega} \sigma_n \leq \tau_1$.
To show $\tau_0 = \tau_1$, it suffices to show $\tau_1 \leq \tau_0$, so it suffices to show $L_{\tau_0}\prec L_{\omega_1}$.
Since for all $\alpha$, $L_{\alpha}\prec_{\Sigma_{n+1}} L_{\omega_1}$ implies $L_{\alpha}\prec_{\Sigma_n}L_{\omega_1}$, we have $\sigma_n\leq \sigma_{n+1}$ for all $n$, so $L_{\sigma_n}\subseteq L_{\sigma_{n+1}}$ for all $n$. Now either $\tau_0$ is a limit ordinal, or $\tau_0 = \sigma_n$ for some $n$. In either case, $L_{\tau_0} = \bigcup_{n\in \omega} L_{\sigma_n}$.
Hence $L_{\tau_0}\prec L_{\omega_1}$ follows from the following general claim.
Claim: For any transitive set $N$ and any chain of transitive subsets $(M_n)_{n\in \omega}$ with $M_n\subseteq M_{n+1}$ and $M_n\preceq_{\Sigma_n} N$ for all $n$, we have $M_\omega = \bigcup_{n\in \omega} M_n \preceq N$.
Proof: Step 1: For all $m \geq n$, $M_m\preceq_{\Sigma_n} M_{m+1}$. Indeed, if $\varphi(\overline{x})$ is a $\Sigma_n$ formula and $\overline{a}\in M_m$, we have $M_m\models \varphi(\overline{a})$ iff $N\models \varphi(\overline{a})$ iff $M_{m+1}\models \varphi(\overline{a})$, since $M_m\preceq_{\Sigma_n} N$ and $M_{m+1}\preceq_{\Sigma_n} N$.
Step 2: For all $m\geq n$, $M_m\preceq_{\Sigma_n} M_\omega$. This is because $M_\omega$ is the union of the chain $(M_m)_{m\geq n}$, which is a $\Sigma_n$-elementary chain by Step 1. But for completeness, let's expand the proof, which is by induction on $n$.
In the base case, $M_m\preceq_{\Sigma_0} M_\omega$ for all $m$, since $M_m$ is a transitive subset of $M_\omega$. Now suppose $M_m\preceq_{\Sigma_n} M_\omega$ for all $m\geq n$. We will show $M_{m}\preceq_{\Sigma_{n+1}} M_\omega$ for all $m\geq n+1$. So fix $m\geq n+1$. We can write an arbitrary $\Sigma_{n+1}$ formula as $\exists \overline{x}\,\varphi(\overline{x},\overline{y})$, with $\varphi$ a $\Sigma_n$ formula. Fix $\overline{b}\in M_m$.
If $M_m\models \exists \overline{x}\,\varphi(\overline{x},\overline{b})$, then there is some $\overline{a}\in M_m$ such that $M_m\models \varphi(\overline{a},\overline{b})$. Since $M_m\preceq_{\Sigma_n} M_\omega$ by induction, $M_\omega\models \varphi(\overline{a},\overline{b})$, so $M_\omega\models \exists \overline{x}\,\varphi(\overline{x},\overline{b})$.
Conversely, if $M_\omega\models \exists \overline{x}\,\varphi(\overline{x},\overline{b})$, then there is some $\overline{a}\in M_\omega$ such that $M_\omega\models \varphi(\overline{a},\overline{b})$. Pick $m' \geq m$ large enough so that $\overline{a}\in M_{m'}$. Since $m'\geq n$, by induction $M_{m'}\preceq_{\Sigma_n} M_\omega$, so $M_{m'}\models \varphi(\overline{a},\overline{b})$, and hence $M_{m'}\models \exists \overline{x}\,\varphi(\overline{x},\overline{b})$. But now $M_m\preceq_{\Sigma_{n+1}} M_{m'}$ by Step 1, so $M_{m}\models \exists \overline{x}\,\varphi(\overline{x},\overline{b})$. This concludes Step 2.
Step 3: Finally, to show $M_\omega\preceq N$, we apply the Tarski-Vaught test. Consider a formula $\varphi(x,\overline{y})$ and a tuple $\overline{b}$ from $M_\omega$ such that $N\models \exists x\, \varphi(x,\overline{b})$. Let $n$ be such that $\exists x\, \varphi(x,\overline{y})$ is $\Sigma_n$. If $\overline{b} = (b_1,\dots,b_k)$, for each $1\leq i \leq k$, let $n_i$ be such that $b_i\in M_{n_i}$. Now let $N = \max(n,n_1,\dots,n_k)$. We have $\overline{b}\in M_N$ and $M_N\preceq_{\Sigma_n} N$, so $M_N\models \exists x\, \varphi(x,\overline{b})$. Now since $n\leq N$, we have $M_N\preceq_{\Sigma_n} M_\omega$ by Step 2, so also $M_\omega\models \exists x\, \varphi(x,\overline{b})$. Thus $M_\omega \preceq N$.