Absolute and Square Summability & Integration of sinc function
Neither the integral

$$I_1=\int_{0}^{\infty}\left|\frac{\sin(x)}{x}\right|dx\tag{1}$$

nor the sum

$$S_1=\sum_{n=1}^{\infty}\left|\frac{\sin(n)}{n}\right|\tag{2}$$

converge. So the sinc function (sequence) is not in $L_1$ ($l_1$). A proof of the fact that $I_1$ diverges can be found here. For the sum $S_1$, note that since $|\sin(n)|\ge \sin^2(n)$ (for $n\in\mathbb{R}$) we have

$$S_1\ge \sum_{n=1}^{\infty}\frac{\sin^2(n)}{n}\tag{3}$$

Here it is shown that the series on the right-hand side of $(3)$ diverges, so $S_1$ must also diverge.

The integral

$$I_2=\int_{0}^{\infty}\left(\frac{\sin(x)}{x}\right)^2dx=\frac{\pi}{2}\tag{4}$$

can be computed in many different ways. Two possibilities are shown in the answers to this question.

Also note that

$$\int_{0}^{\infty}\left(\frac{\sin(x)}{x}\right)^2dx=\int_{0}^{\infty}\frac{\sin(x)}{x}dx\tag{5}$$

The computation of the sum

$$S_2=\sum_{n=1}^{\infty}\left(\frac{\sin(n)}{n}\right)^2=\frac{\pi-1}{2}\tag{6}$$

is discussed in this question and its answers. Note that a relation similar to $(5)$ holds:

$$\sum_{n=1}^{\infty}\left(\frac{\sin(n)}{n}\right)^2=\sum_{n=1}^{\infty}\frac{\sin(n)}{n}\tag{7}$$

So clearly, the sinc function (sequence) is in $L_2$ ($l_2$).