Does a bounded function like this exist? is it possible to find a function $f$ such that $f \in L^{\infty}(\mathbb{T}^1)$ and $$\sum_{j=0}^{+\infty} \left(\sum_{n \in \mathbb{Z}: 2^j \le |n|<2^{j+1}}|\widehat{f}_{n}|^{2}  \right)^{1/2}=+\infty  \hspace{0.5cm}\text{?}$$
 A: Consider the products $p_n(x)=\Pi_{j=1}^n(1+i\frac{\cos 4^j x}{j})=1+\sum_{1\le k\le m_n}a_k\cos kx$ where $m_n=4+..+4^n$
Noting that $m_n < 4^{n+1}-4^n-4^{n-1}-...-4$ it follows that when we pass from $p_n$ to $p_{n+1}$ we do not change the $a_k$ from $p_n$ so we have an unambiguous definition of $f(x)=1+\sum_{k \ge 1}a_k\cos kx$ for the infinite product above when $n\to \infty$.
In the same vein note that each $k \ge 1$ with possible non zero coefficient appears only once as a finite combination of the type $4^n \pm 4^{n_1} \pm ...$ with $n>n_1>..$ so in particular $a_{4^n}=i/n, n \ge 1$.
Since $\sum 1/j^2$ finite it is easy to see that $\sum |a_k|^2$ finite so $f$ defines a Fourier series.
Now $|p_n(x)|^2 \le \Pi_{j\ge 1}(1+1/j^2)<\infty$ so the partial sums of $f$ up to index $m_n$ are uniformly  bounded, hence by general Fourier theory it follows that $f$ is bounded.
However $|\hat f_{4^n}|=|a_{4^n}/2|=1/(2n), n \ge 1$ so the inner sum in the OP for $j=2n$ is at least $1/(2n)$ so indeed the full sum is infinite.
