For $f\in L^1(\mathbb{R}^n)$, let $P_{2^k}f,k\in\mathbb{Z}$ be the inverse Fourier transform of $1_{2^k\le|\xi|<2^{k+1}}\hat{f}(\xi)$.
Then, do we have the following?
$$\sum_{k\in\mathbb{Z}}\|P_{2^k}f\|_{L^1}^2\lesssim\|f\|_{L^1}^2$$
Here, $A\lesssim B$ means $A\le CB$ for some fixed constant $C<\infty$.
(The notation $P_k$ makes more sense here than $P_{2^k}$.)
It's honestly obvious that the answer is no, because in fact $P_kf$ is not typically even an $L^1$ function, since its Fourier transform is not continuous.
Which is why you might want to consider a different $P_kf$, replacing the charactristic function of that annulus by a smooth cutoff function. If you do that the answer is not so obvious; I still think it's very likely the answer is no, just because these things never work for $p=1$.
Edit: No wait, it's actually very easy to see the revised version also fails. If true it would imply $$\sum|\hat f(2^kx)|^2<\infty\quad(f\in L^1,x\in\Bbb R^n),$$which we know is not so, since the Riemann-Lebesgue Lemma is best possible (If $\psi>0$ and $\psi\to0$ at $\infty$ then there exists $f\in L^1$ with $\hat f\ge\psi$.)