$\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}$This is not quite obvious, and it has hardly anything to do with the reverse Fatou lemma.

Indeed, for all $s\in[0,t]$, let
\begin{equation*}
l_n(s):=\sup_{m\colon m\ge n}\ell^m(s),
\end{equation*}
so that
\begin{equation*}
\ell^n(s)\le l_n(s)\downarrow l(s):=\limsup_n\ell^n(s). \tag{0}
\end{equation*}
So,
\begin{align*}
&\limsup_n P(\exists s\in[0,t]\ x+B_s\le\ell^n(s)) \\
\le &\limsup_n P(\exists s\in[0,t]\ x+B_s\le l_n(s)) \\
=&\lim_n P(\exists s\in[0,t]\ x+B_s\le l_n(s))=P(A),
\end{align*}
where
\begin{align*}
A&:=\{\forall n\ \exists s\in[0,t]\ x+B_s\le l_n(s)\} \\
& =\{\forall n\ge m\ \exists s\in[0,t]\ x+B_s\le l_n(s)\};
\end{align*}
here and in what follows, $m$ is any natural number.
So, it is enough to show that
\begin{equation*}
P(A)\overset{\text{(?)}}\le P(C), \tag{1}
\end{equation*}
where
\begin{equation*}
C:=\{\exists s\in[0,t]\ x+B_s\le l(s)\}.
\end{equation*}
We shall actually show that
\begin{equation*}
P(A\setminus C)\overset{\text{(?)}}=0, \tag{2}
\end{equation*}
which will of course imply (1).

Suppose that event $A$ occurs. For all $n$, let

\begin{equation*}
s_n:=\inf\{s\in[0,t]\colon x+B_s\le l_n(s)\};
\end{equation*}
of course, $s_n$ is a random variable (r.v., with values in $[0,t]$ on $A$), depending on the random path of the Brownian motion $(B_t)$.
Moreover, since $l_n(s)\downarrow l(s)$ for all $s\in[0,t]$, we have
\begin{equation*}
s_n\uparrow s_*
\end{equation*}
for some r.v. $s_*$, with values in $[0,t]$ on $A$.

Consider first the case when $A$ occurs and $s_n<s_*$ for all $n$. Then for all $n$ there is some $t_n\in[s_n,s_*)$ such that $x+B_{t_n}\le l_n(t_n)$.
Also, $l_n(s)$ is nondecreasing in $s\in[0,t]$. So, for all $n$, we have $x+B_{t_n}\le l_n(s_*)$.
So,
\begin{equation*}
x+B_{s_*}=\lim_n(x+B_{t_n})\le\lim_n l_n(s_*)=l(s_*).
\end{equation*}
Thus,
\begin{equation*}
A\cap\{\forall n\ s_n<s_*\}\subseteq C. \tag{2.5}
\end{equation*}

If $A$ occurs and $s_n=s_*=t$ for some $n$ (and hence for all large enough $n$), then for such $n$ we have $x+B_t\le l_n(t)$ and hence $x+B_t\le l(t)$. Thus,
\begin{equation*}
A\cap\{\exists n\ s_n=s_*=t\}\subseteq C. \tag{2.75}
\end{equation*}

If $A$ occurs and if $s_n=s_*<t$ for some $n$ (and hence for all large enough $n$) and if $s_*$ is a point of discontinuity of the function $l$, then for large enough $n$ we have $x+B_{s_*}=x+B_{s_n}\le l_n(s_*+)$, so that $x+B_{s_*}\le l^+(s_*)$, where $l^+(s):=\lim_n l_n(s+)$.
So,
\begin{equation*}
x+B_d\le l^+(d) \tag{3}
\end{equation*}
at some point $d\in D$, where $D$ is the set of all points of discontinuity of the nondecreasing function $l$; there are at most countably many such points.

Note that for all $u\in[0,t)$ and all $s\in(u,t]$ we have $l(s)=\lim_n l_n(s)\ge\lim_n l_n(u+)=l^+(u)$.
So, $l(s)\ge l^+(u)$ for all $u\in[0,t)$ and all $s\in(u,t]$.
Now suppose that $C$ does not occur, so that $x+B_s>l(s)$ for all $s\in[0,t]$ and hence $x+B_s> l^+(d)$ for each $d\in D$ and all $s\in(d,t]$. In view of, say, the (local) law of the iterated logarithm for the Brownian motion, for each $d\in[0,t)$ the event
$\{x+B_d\le l^+(d),\ x+B_s>l^+(d)\ \forall s\in(d,t]\}$ has the zero probability.
In view of (3) and because the set $D$ is at most countable, we conclude that
\begin{equation*}
P(A\cap\{\exists n\ s_n=s_*<t,s_*\in D\}\setminus C)=0. \tag{4}
\end{equation*}

Suppose finally that $A$ occurs and $s_n=s_*<t$ for some $n$ (and hence for all large enough $n$) and $s_*$ is a point of continuity of the function $l$. Take now any real $\ep>0$. Then there is some real $\de>0$ such that $l(s_*+\de)\le l(s_*)+\ep$. So, for all large enough $n$,
\begin{equation*}
x+B_{s_*}=x+B_{s_n}\le l_n(s_n+\de)=l_n(s_*+\de)\to l(s_*+\de)\le l(s_*)+\ep.
\end{equation*}
Letting now $\ep\downarrow0$, we get $x+B_{s_*}\le l(s_*)$. So,
\begin{equation*}
A\cap\{\exists n\ s_n=s_*<t,s_*\notin D\}\subseteq C. \tag{5}
\end{equation*}

Collecting (2.5), (2.75), (4), and (5), we confirm (2), as desired.