$\newcommand\R{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}\newcommand{\vpi}{\varphi}$For $\phi=c\,1_{[-A,\infty)}$ with $c\ge0$ and $A\ge0$, the expectation in question converges to
\begin{equation*}
\frac1{1+(e^c-1)e^{A\sqrt{\ln4}}}. \tag{1}\label{1}
\end{equation*}
Indeed, let $n:=2^N\to\infty$. Let
\begin{equation*}
h:=\phi=c\,1_{[-A,\infty)} \tag{2}\label{2}
\end{equation*}
for some real with $c\ge0$ and $A\ge0$.
Let $V_i:=X_i-\max_{1\le j\le n}X_j$. Let $g\colon\R^n\to\R$ be
any nonnegative Borel-measurable function that is symmetric (with respect to any permutation of its arguments). Let $f$ denote the pdf of each $X_j$, so that
\begin{equation*}
f(x)=\frac1s\,\vpi\Big(\frac xs\Big)
\end{equation*}
for all real $x$, where $\vpi$ is the standard normal pdf and
\begin{equation*}
s:=\sqrt N=\sqrt{\log_2 n}=\sqrt{\frac{\ln n}{\ln2}}. \tag{3}\label{3}
\end{equation*}
Then
\begin{equation*}
\begin{aligned}
&Eg(V_1,\dots,V_n) \\
&=n\,Eg(X_1-X_1,X_2-X_1,\dots,X_n-X_1)\,1(X_1>\max_{2\le j\le n}X_j) \\
&=n\,\int_{\R^n}g(0,x_2-x_1,\dots,x_n-x_1)\,1(x_1>\max_{2\le j\le n}x_j)
\prod_{j=1}^n f(x_j)dx_j\, \\
&=n\,\int_\R f(x_1)\,dx_1\,\int_{(-\infty,0)^n}g(0,v_2,\dots,v_n)\,
\prod_{j=2}^n f(x_1+v_j)dv_j.
\end{aligned}
\end{equation*}
Then the expectation in question is
\begin{equation*}
L:=L_n:=ne^{-c}I, \tag{3.5}\label{3.5}
\end{equation*}
where
\begin{equation*}
\begin{aligned}
I:=I_n&:=\int_\R f(x_1)\,dx_1\,\Big(\int_{-\infty}^0 e^{-h(v)} f(x_1+v)\,dv\Big)^{n-1} \\
&=\int_\R dz\,\vpi(z)\,H(z)^{n-1},
\end{aligned}
\end{equation*}
where
\begin{equation*}
\begin{aligned}
H(z)&:=\int_{-\infty}^z e^{-h(s(w-z))}\vpi(w)\,dw \\
&=e^{-c}\Phi(z)+(1-e^{-c})\Phi(z-A/s),
\end{aligned}
\end{equation*}
where $\Phi$ is the standard normal cdf (and $h$ and $s$ are as defined in \eqref{2} and \eqref{3}).
Let
\begin{equation*}
z_\ep:=\sqrt{(2-\ep)\ln n},
\end{equation*}
where $\ep\in(0,2)$.
Note that
\begin{equation*}
I=J_1+J_2+J_3, \tag{4}\label{4}
\end{equation*}
where
\begin{equation*}
J_1:=\int_{-\infty}^{z_\ep} dz\,\vpi(z)\,H(z)^{n-1},\quad
J_2:=\int_{z_\ep}^{z_0} dz\,\vpi(z)\,H(z)^{n-1},\quad
J_3:=\int_{z_0}^\infty dz\,\vpi(z)\,H(z)^{n-1}.
\end{equation*}
Noting that $0<H\le\Phi<1$ and letting
\begin{equation*}
G:=1-\Phi,
\end{equation*}
we see that, for each $\ep\in(0,2)$,
\begin{equation*}
nJ_1\le\int_{-\infty}^{z_\ep} dz\,\vpi(z)\,n\Phi(z)^{n-1}
=\Phi(z_\ep)^n \\
=(1-G(z_\ep))^n\le\exp\{-nG(z_\ep)\}=o(1/n), \tag{4.5}\label{4.5}
\end{equation*}
since
\begin{equation*}
nG(z_\ep)=n\exp\Big\{-\frac{z_\ep^2}{2+o(1)}\Big\}
=n\exp\Big\{-\frac{2-\ep}{2+o(1)}\,\ln n\Big\}=n^{\ep/(2+o(1))}.
\end{equation*}
So, the conclusion
\begin{equation*}
nJ_1=o(1/n) \tag{5}\label{5}
\end{equation*}
will hold if $\ep=\ep_n$, for some sequence $\ep_n\downarrow0$. In what follows, it is indeed assumed that $\ep=\ep_n\downarrow0$.
Next,
\begin{equation*}
J_3\le\int_{z_0}^\infty dz\,\vpi(z)=G(z_0)<\frac{\vpi(z_0)}{z_0}=o(\vpi(z_0))=o(1/n).
\tag{6}\label{6}
\end{equation*}
Further,
\begin{equation*}
\begin{aligned}
H'(z)&=e^{-c}\vpi(z)+(1-e^{-c})\vpi(z-A/s) \\
&=\vpi(z)[e^{-c}+(1-e^{-c})e^{Az/s}e^{-A^2/(2s^2)}] \\
&\sim\vpi(z)[e^{-c}+(1-e^{-c})e^{Az/s}],
\end{aligned}
\end{equation*}
since $s\to\infty$.
Also, for $z\in[z_\ep,z_0]$ we have $z\sim z_0$ (since $\ep\downarrow0$) and hence
\begin{equation*}
z/s\to z_0/s=\sqrt{\ln4}.
\end{equation*}
So,
\begin{equation*}
\begin{aligned}
J_2&=\int_{z_\ep}^{z_0} dz\,\frac{\vpi(z)}{H'(z)}\,H'(z)H(z)^{n-1} \\
&\sim\frac1{e^{-c}+(1-e^{-c})e^{A\sqrt{\ln4}}}\,\int_{z_\ep}^{z_0} dz\,H'(z)H(z)^{n-1}.
\end{aligned}
\tag{7}\label{7}
\end{equation*}
Also,
\begin{equation*}
\begin{aligned}
n\int_{-\infty}^{z_\ep} dz\,H'(z)H(z)^{n-1}= H(z_\ep)^n\le\Phi(z_\ep)^n =o(1),
\end{aligned}
\tag{8}\label{8}
\end{equation*}
as shown in \eqref{4.5}.
Also,
\begin{equation*}
\begin{aligned}
\int_{z_0}^\infty dz\,H'(z)H(z)^{n-1}
&\le\int_{z_0}^\infty dz\,H'(z) \\
&=e^{-c}G(z_0)+(1-e^{-c})G(z_0-A/s) \\
&\le G(z_0-A/s) \\
&\le \frac{\vpi(z_0-A/s)}{z_0-A/s} \\
&=O\Big(\frac{\vpi(z_0)}{z_0-A/s}\Big)
=o(\vpi(z_0))=o(1/n).
\end{aligned}
\tag{9}\label{9}
\end{equation*}
Collecting \eqref{7}, \eqref{8}, and \eqref{9}, we get
\begin{equation*}
\begin{aligned}
J_2&\sim\frac1{e^{-c}+(1-e^{-c})e^{A\sqrt{\ln4}}}\, \\
&\times\Big(\int_{-\infty}^\infty dz\,H'(z)H(z)^{n-1} \\
&\quad-\int_{-\infty}^{z_\ep} dz\,H'(z)H(z)^{n-1}-\int_{z_0}^\infty dz\,H'(z)H(z)^{n-1}\Big) \\
&=\frac1{e^{-c}+(1-e^{-c})e^{A\sqrt{\ln4}}}\,(1/n-o(1/n)-o(1/n)).
\end{aligned}
\tag{10}\label{10}
\end{equation*}
Finally, collecting \eqref{3.5}, \eqref{4}, \eqref{5}, \eqref{6}, and \eqref{10}, we get the limit \eqref{1} for $L$.
It follows that $\sum_{k=1}^n h(X_k-\max_{1\le i\le n} X_i)$ converges in distribution to a geometrically distributed random variable $Y$ such that
\begin{equation}
P(Y=j)=\tfrac1B\,(1-\tfrac1B)^{j-1}\,1(j\in\{1,2,\dots\}),
\end{equation}
where
\begin{equation}
B:=e^{A\sqrt{\ln4}}>1.
\end{equation}
This result could probably be obtained directly, without using the Laplace transform.
