$N(p)$ can grow to $\infty$ however fast. Indeed, let $g\colon(0,\infty)\to(1,\infty)$ be any (say) strictly increasing continuous function, with $g(\infty-)=\infty$. 
We want to show that for some nonnegative function $f$ such that $f\in L^p$ for all $p>0$ we have  
\begin{equation*}
	N(p)\ge g(p).  \tag{1}\label{1}
\end{equation*}

Note that  
\begin{equation*}
	G(p):=g(p)^p>1
\end{equation*}
is also strictly increasing and continuous in $p$. Let 
\begin{equation*}
	h_p:=\frac1{G(p)}\in(0,1)
\end{equation*}
and 
\begin{equation*}
	a(x):=\frac2{G^{-1}(1/x)};
\end{equation*}
everywhere here, by default,  $x\in(0,1)$. 
Note that $a(x)\downarrow0$ as $x\downarrow0$. 
Moreover, the condition that $g$ is increasing means exactly that  
\begin{equation*}
	b(x):=a(x)\ln\tfrac1x
\end{equation*}
is decreasing in $x$. 

Letting now 
\begin{equation*}
	f(x):=x^{-a(x)}=e^{b(x)} 
\end{equation*}
and recalling that $a(x)\downarrow0$ as $x\downarrow0$, for each real $p>0$ we have $f(x)^p\le x^{-1/2}$ for all $x$ in a right neighborhood of $0$. Also, $f$ is bounded outside any right neighborhood of $0$. So, $f\in L^p$ for all $p>0$. 

On the other hand, recalling that $b$ is decreasing, we have 
\begin{equation*}
	N(p)^p\ge \int_0^{h_p} f^p
	=\int_0^{h_p} e^{pb(x)}\,dx
	\ge h_p e^{pb(h_p)}=1/h_p=G(p)=g(p)^p,
\end{equation*}
so that \eqref{1} follows, as desired.