$\newcommand{\de}{\delta}\newcommand{\al}{\alpha}\newcommand{\be}{\beta}\newcommand{\R}{\mathbb R}\newcommand{\vpi}{\varphi} $Without loss of generality, the variance of $X_1$ is $1$.
Let $H:=\mathrm H$, which let us assume to denote the differential entropy, so that $H(X)=\int_{\mathbb R} f(x)\ln\frac1{f(x)}\,dx$ for a random variable $X$ with pdf $f$.
If the pdf of $X_1$ is bounded or, more generally, the pdf of $\bar X_k$ is bounded for some natural $k$, then, by a local limit theorem (see e.g. Theorem 7 in Section 2 of Chapter VII), the pdf (say $f_n$) of $\sqrt n\,\bar X_n$ converges uniformly on $\R$ to the standard normal pdf (say $\vpi$) as $n\to\infty$. Note that $\vpi\le1/\sqrt{2\pi}<1$. So, for all large enough $n$ we have $f_n\le1$ and hence $f_n\ln\frac1{f_n}\ge0$.
So, by Fatou's lemma,
\begin{equation*}
\liminf_{n\to\infty}H(\sqrt n\,\bar X_n)=\liminf_{n\to\infty}\int_\R f_n\ln\frac1{f_n}
\ge\int_\R \vpi\ln\frac1\vpi=H(Z)
\end{equation*}
if $Z\sim N(0,1)$.
On the other hand, $H(\sqrt n\,\bar X_n)\le H(Z)$ for all $n$, since the variance of $\sqrt n\,\bar X_n$ is $1$ and $Z$ maximizes the differential entropy among all absolutely continuous random variables with variance $1$.
So,
\begin{equation*}
\lim_{n\to\infty}H(\sqrt n\,\bar X_n)=H(Z). \tag{1}\label{1}
\end{equation*}
Without the condition that the pdf of $\bar X_k$ be bounded for some natural $k$, the conclusion \eqref{1} can fail to hold, in a rather dramatic manner.
An example when $H(\sqrt n\,\bar X_n)=-\infty$ for all $n$, so that \eqref{1} fails to hold, is given by the formula
\begin{equation*}
f=\frac1s\,\sum_{k\ge1}c_k\,1_{[x_k,3x_k/2]}
\end{equation*}
for the pdf of $X_1$, where $s:=\sum_{k\ge1}c_k x_k/2=\pi^2/12$,
$c_k:=e^{2^k}/k^2$ and $x_k:=e^{-2^k}$; of course, in this example the pdf of $\bar X_k$ is not bounded for any natural $k$.
The main idea here is that the pdf (say $f_n$) of $Y_n:=\sqrt n\,\bar X_n$ may remain very large in (say) a neighborhood of $0$, so much so that $H(Y_n)=\int f_n\ln\frac1{f_n}=-\infty$.
Details on this example: Let $\de_k:=x_k/2$. Consider the class $G$ of all pdf's $g$ such that
\begin{equation*}
g\ge\sum_{j\ge1}a_j\,1_{[u_j,u_j+\de_j]}, \tag{2}\label{2}
\end{equation*}
where
\begin{equation*}
a_j:=\frac{C_1}{j^p\,\de_j}\quad\text{and}\quad u_j:=\frac{\al x_j}2 \tag{3}\label{3}
\end{equation*}
for some real $C_1>0$, some real $p>0$, some real $\al\ge1$, and all $j\ge1$.
Clearly, \eqref{1} implies that $f\in G$. Moreover, the pdf's in the class $G$ are very lacunary, which allows us to control the convolutions of any two pdf's in $G$.
More specifically, let us show that -- crucially -- the class $G$ is closed w.r. to the convolution:
Take any $g\in G$, so that \eqref{2} and \eqref{3} hold, and then take any $h\in G$, so that
\begin{equation*}
h\ge\sum_{j\ge1}b_j\,1_{[v_j,v_j+\de_j]}, %\tag{2a}\label{2a}
\end{equation*}
where
\begin{equation*}
b_j:=\frac{C_2}{j^q\,\de_j}\quad\text{and}\quad v_j:=\frac{\be x_j}2 %\tag{3a}\label{3a}
\end{equation*}
for some real $C_2>0$, some real $q>0$, some real $\be\ge1$, and all $j\ge1$.
Note that $1_{[u_j,u_j+\de_j]}*1_{[v_j,v_j+\de_j]}\ge\frac12\,1_{[w_j,w_j+\de_j]}$, where $w_j:=\frac{(\al+\be+1) x_j}2$. So,
\begin{equation*}
g*h\ge\sum_{j\ge1}d_j\,1_{[w_j,w_j+\de_j]},
\end{equation*}
where $d_j:=a_jb_j\de_j=\frac{C_1C_2/2}{j^{p+q}\de_j}$. Thus, indeed the class $G$ is closed w.r. to the convolution.
So, to complete the consideration of the example, it remains to show that
\begin{equation*}
H(g):=-\int_\R g\ln g=-\infty \tag{4}\label{4}
\end{equation*}
for any $g$ such that \eqref{2} and \eqref{3} hold. Take indeed any such $g$. Clearly, for some natural $j_p$ and all natural $j\ge j_p$ we have $a_j\ge1$ and hence $g\ge1$ on the interval $[u_j,u_j+\de_j]$. Also, the intervals $[u_j,u_j+\de_j]$ are pairwise disjoint.
Also, $t\ln t$ is increasing in $t\ge1$.
It follows that
\begin{equation*}
\int_\R g\ln g\,1(g\ge1)\ge \sum_{j\ge j_p}a_j\,\de_j\,\ln a_j
=\sum_{j\ge j_p}\frac{C_1}{j^p}\,\ln\frac{2C_1}{j^p\,e^{-2^j}}=\infty. \tag{5}\label{5}
\end{equation*}
On the other hand, $t\ln t\ge-1/e$ for all real $t>0$. So,
\begin{equation*}
\int_\R g\ln g\,1(g<1)\ge-\frac1e\, \int_{[0,u_1+\de_1]}1>-\infty. \tag{6}\label{6}
\end{equation*}
Now \eqref{4} follows from \eqref{5} and \eqref{6}. $\quad\Box$
