Let the underlying space be $X = (0,1)$ with the usual Lebesgue measure. For convenience choose $p = 1$.
Let
$$ f_k = k^{1 - \alpha(k)}\mathbf{1}_{(0,1/k)} $$
where $\mathbf{1}_A$ is the indicator function of the set $A$. If $\alpha$ satisfies:
- $\alpha$ is positive
- $\alpha$ is monotonically decreasing in $k$, with limit 0 as $k\to\infty$
- $\lim_{k \to \infty} k^{-\alpha(k)} = 0$
Then 2. implies that for any $q > 1$, $\|f_k\|_q \nearrow +\infty$. And 3. implies that for $p = 1$, $\|f_k\|_1 \searrow 0$.
A possible choice of $\alpha$ is $1 / \log\log k$. Since we have
$$ \log (k^{-\alpha(k)}) = - \frac{\log k}{\log \log k} $$
with limit $-\infty$, this implies the desired result 3.