I just solved this problem while self studying Rudin and was looking for answers to compare to mine. I will post my original solution below for the sake of those who have bookmarked the question and/or might benefit from a different solution.
$\textbf{Disclaimer}$
My strategy involves bounding the measures of useful disjoint subsets, the union of which equals $(0,1)$, rather than building a specific function. I am not sure if there is a more obvious or elegant solution. Also:
I have not had this solution examined by anybody other than myself so please if any mistakes or ambiguities are spotted let me know.
$\textbf{Solution}$
Fix some $\Phi:(0,\infty)\rightarrow(0,\infty)$ with $\Phi(x)\rightarrow\infty$ as $x\rightarrow\infty$. Assume without loss of generality that $\Phi\geq1$, that $\Phi$ is nondecreasing, and let $f\geq0$ be (Lebesgue) measurable. Define
\begin{equation}\tag{1}
E_n=f^{-1}[n,n+1),\qquad n\in\mathbb{N}.
\end{equation}
Since $E_n$ is clearly the union of two measurable sets, $E_n$ is, for all $n$, measurable. Moreover, for all $p\geq1$ we have
\begin{equation}\tag{2}
\left(\int_0^1f^p\ dm\right)^\frac{1}{p}<(\sum_{n=0}^\infty(n+1)^pm(E_n))^\frac{1}{p}\leq\sum_{n=0}^\infty(n+1)m(E_n)^{1/p}.
\end{equation}
Associate to each $n\in\mathbb{N}$ some $p_n>0$ with $\Phi(p_n)\geq(n+1)2^{n+1}$. Since for all $n$ it holds that $m(E_n)\leq1$, and since $\Phi$ is nondecreasing, we obtain
\begin{equation}\tag{3}
m(E_n)^{1/p}\leq\frac{\Phi(p)}{(n+1)2^{n+1}},\qquad\forall p\geq p_n.
\end{equation}
The hypothesis $\Phi(p)\geq1,\ \forall p$ gives
\begin{equation}\tag{4}
\left(\frac{\Phi(p)}{(n+1)2^{n+1}}\right)^p\geq\left(\frac{1}{(n+1)2^{n+1}}\right)^p,\qquad\forall n\in\mathbb{N},\ \forall p.
\end{equation}
But the infimum over all $p\in[0,p_n)$ of the right side of (4) is equal to this very term in the case $p=p_n$. Hence (4) gives
\begin{equation}\tag{5}
\inf\left\{\left(\frac{\Phi(p)}{(n+1)2^{n+1}}\right)^p:0\leq p<p_n\right\}\geq\left(\frac{1}{(n+1)2^{n+1}}\right)^{p_n}.
\end{equation}
According to (5), (3) and (2), choosing $m(E_n)$ (and $f$) with
\begin{equation}\tag{6}
0<m(E_n)\leq\left(\frac{1}{(n+1)2^{n+1}}\right)^{p_n},\ \forall n\in\mathbb{N}
\end{equation}
will ensure that
\begin{equation}\tag{7}
||f||_p<\sum_{n=0}^\infty(n+1)m(E_n)^{1/p}\leq\sum_{n=0}^\infty\frac{\Phi(p)}{2^{n+1}}=\Phi(p)
\end{equation}
for all $p\geq 1$, while (6) and (1) give $||f||_\infty=\infty$. Hence $||f||_p$ can indeed tend to $\infty$ arbitrarily slowly. $\qquad\blacksquare$
