Let $$a_n:=\dfrac23(4^n + 2).$$
By the triangle inequality, with probability $1$ we have 
$|S_{n-1}|\le\sum_{k=1}^{n-1}|X_k|=\sum_{k=1}^{n-1}4^k=4^n-a_n$; that is, $P(|S_{n-1}|\le4^n-a_n)=1$. The equality $X_n=S_n-S_{n-1}$ and the triangle inequality imply that on the event $\{|S_n|<a_n,|S_{n-1}|\le4^n-a_n\}$ we have 
$$|X_n|\le|S_n|+|S_{n-1}|<a_n+(4^n-a_n)=4^n$$
and hence $|X_n|<4^n$. So, on the event $\{|X_n|=4^n,|S_{n-1}|\le4^n-a_n\}$ we have $|S_n|\ge a_n$. So, 
$$P(|S_n|\ge a_n)\ge P(|X_n|=4^n,|S_{n-1}|\le4^n-a_n)=1,$$
because $P(|X_n|=4^n)=1$ and $P(|S_{n-1}|\le4^n-a_n)=1$. 
Hence, 
$$P(|S_n|\ge a_n)=1. \tag{1}\label{1}$$
On the other hand,
$$P(|S_n|=a_n)\ge P(X_n=4^n,S_{n-1}=-(4^n-a_n)) \\
=P(X_n=4^n,X_{n-1}=-4^{n-1},\dots,X_1=-4^1)=2^{-n}>0$$
and hence for all $r>a_n$ we have 
$$P(|S_n|\ge r)\le P(|S_n|>a_n)\le1-2^{-n}<1. \tag{2}\label{2}$$
So, 
$$A_n=a_n.\quad\Box$$

---

**Details on the conclusion $A_n=a_n$, in response to a comment by the OP:** We have 
$$A_n=\sup E,\quad\text{where}\quad 
E:=\{r\in\Bbb R\colon P(|S_n|\ge r)=1\}. $$
By \eqref{2}, for any $r>a_n$ we have $r\notin E$; that is, $E\subseteq(-\infty,a_n]$. Also, by \eqref{1}, $a_n\in E$. We conclude that $\sup E=\max E=a_n$; that is, $A_n=a_n$.