For $p\geqslant 1$ there is some $f\in L^1$ s.t. $\sum_{k\in \Bbb Z }|\hat f(k)|^p=\infty $

If for $0 < \alpha < 1$ you let $f(t) = |t|^{\alpha - 1}$ then a direct calculation shows that for some constant $c$ one has $\hat{f}(k) = c|k|^{-\alpha} + o(|k|^{-\alpha})$ for $k > 0$ and $\hat{f}(k) = \bar{c}|k|^{-\alpha} + o(|k|^{-\alpha})$ for $k < 0$. Thus taking $\alpha$ close enough to $0$ will give an example for any given $p$.

Some details on the calculation: by definition one has $$\hat{f}(k) = {1 \over 2\pi} \int_{-\pi}^{\pi}|t|^{\alpha - 1}e^{-ikt} dt$$ Say $k > 0$. Then changing variables to $u = tk$ this equals $${1 \over 2\pi} k^{-\alpha}\int_{-k\pi}^{k\pi}|u|^{\alpha - 1}e^{-iu} du$$ Since the improper integral $\int_{-\infty}^{\infty}|u|^{\alpha - 1}e^{-iu} du$ converges to some limit $L$ this equals $${L \over 2\pi} k^{-\alpha} - k^{-\alpha}{1 \over 2\pi} \int_{|u| > 2k\pi}|u|^{\alpha - 1}e^{-iu} du$$ Again using that the improper integral converges, the error term is $o(k^{-\alpha})$ as $k \rightarrow \infty$.