$\newcommand{\abs}[1]{\left\vert #1\right\vert}%
\newcommand{\ic}{{\rm i}}$
We didn't calculate the Fourier transform the OP asked for. We just calculate its
derivative. It turns out that it has singularities at $k =0, \pm 1, \pm 2, \pm 4, \pm 6,\ldots$. We believe this is the root of problems which makes hard to evaluates the Fourier transform mentioned above. We hope somebody else can take from our final result.
\begin{align}
\phi\left(k\right)
&\equiv
\int_{\infty}^{\infty}\abs{\sin\left(x\right) \over x}\,{\rm e}^{\ic kx}
\,{\rm d}x
=
2\int_{0}^{\infty}{\abs{\sin\left(x\right)} \over x}\,\cos\left(kx\right)
\,{\rm d}x
\\[3mm]
\phi'\left(k\right)
&=
2\int_{0}^{\infty}\abs{\sin\left(x\right)}\sin\left(kx\right)
\,{\rm d}x\,,
\end{align}
$\abs{\sin\left(x\right)}\quad$ is periodic
$\left(~\mbox{of period}\ \pi~\right)$: $\abs{\sin\left(x\right)}
=
\sum_{n = \infty}^{\infty}A_{n}\cos\left(2nx\right)$ with
$$
\int_{\pi/2}^{\pi/2}\cos\left(2mx\right)\cos\left(2nx\right)\,{\rm d}x
=
{\pi \over 2}\,\delta_{mn}
$$
$$
A_{n}
=
{2 \over \pi}\int_{\pi/2}^{\pi/2}\abs{\sin\left(x\right)}\cos\left(2nx\right)
\,{\rm d}x
=
{4 \over \pi}\,{1 \over 1  4n^{2}}
=
\,{1 \over \pi}\,{1 \over n^{2}  1/4}
$$
Then,
\begin{align}
\phi'\left(k\right)
&=
2\sum_{n = \infty}^{\infty}A_{n}\int_{0}^{\infty}\cos\left(2nx\right)\sin\left(kx\right)
\,{\rm d}x
\\[3mm]&=
\,\Im\sum_{n = \infty}^{\infty}
A_{n}\int_{0}^{\infty}\left[%
{\rm e}^{\ic\left(k  2n\right)x}
+
{\rm e}^{\ic\left(k + 2n\right)x}
\right]\,{\rm d}x
\\[3mm]&=
\,\Im\sum_{n = \infty}^{\infty}A_{n}\left[%
{1 \over \ic\left(k  2n\right)  0^{+}}
+
{1 \over \ic\left(k + 2n\right)  0^{+}}
\right]
\\[3mm]&=
\,\Im\sum_{n = \infty}^{\infty}A_{n}\left(%
{\ic \over 2n  k  \ic 0^{+}}
+
{\ic \over 2n + k + \ic 0^{+}}
\right)
\\[3mm]&=
{1 \over 2}\Re\sum_{n = \infty}^{\infty}A_{n}\left(%
{1 \over n  k/2  \ic 0^{+}}

{1 \over n + k/2 + \ic 0^{+}}
\right)
\\[3mm]&=
\,{1 \over 2\pi}{\cal P}\,k\sum_{n = \infty}^{\infty}
{1 \over n^{2}  1/4}\,{1 \over n^{2}  \left(k/2\right)^{2}}
\\[3mm]&=
\,{1 \over 2\pi}{\cal P}\,\left[%
{16 \over k}
+
2k\sum_{n = 0}^{\infty}
{1 \over n^{2}  1/4}\,{1 \over n^{2}  \left(k/2\right)^{2}}
\right]
\\[3mm]&=
\,{1 \over 2\pi}{\cal P}\,\left\{%
{16 \over k}

{8k \over k^{2}  1}\sum_{n = 0}^{\infty}\left[%
{1 \over n^{2}  1/4}  {1 \over n^{2}  \left(k/2\right)^{2}}
\right]\right\}
\end{align}
\begin{align}
\sum_{n = 0}^{\infty}{1 \over n^{2}  a^{2}}
&=
\sum_{n = 0}^{\infty}{1 \over \left(n + \abs{a}\right)\left(n  \abs{a}\right)}
=
{\Psi\left(\abs{a}\right)  \Psi\left(\abs{a}\right) \over 2\abs{a}}
\\[3mm]&=
{1 \over 2\abs{a}}\left\{%
\Psi\left(\abs{a}\right)  \Psi\left(1 + \abs{a}\right)
+
\pi\cot\left(\pi\left[\abs{a}\right]\right)
\right\}
=
{1 \over 2\abs{a}}\left[%
\,{1 \over \abs{a}}

\pi\cot\left(\pi\abs{a}\right)
\right]
\end{align}
$$
\sum_{n = 0}^{\infty}{1 \over n^{2}  a^{2}}
=
\,{1 \over 2a^{2}}\left[%
1
+
{\pi a \over \tan\left(\pi a\right)}\right]\,,
\qquad
a \not\in {\mathbb Z}
$$
$$
\sum_{n = 0}^{\infty}{1 \over n^{2}  1/4}
=
2\,,
\qquad
\sum_{n = 0}^{\infty}{1 \over n^{2}  \left(k/2\right)^{2}}
=
\,{2 \over k^{2}}\left[1 + {\pi k/2 \over \tan\left(\pi k/2\right)}\right]
$$
\begin{align}
\phi'\left(k\right)
&=
\,{1 \over 2\pi}{\cal P}\,\left[%
{16 \over k}
+
{16k \over k^{2}  1}

{16 \over k}\,{1 \over k^{2}  1}

{16 \over k}\,{1 \over k^{2}  1}\,{\pi k/2 \over \tan\left(\pi k/2\right)}
\right]
\end{align}
\begin{align}
\phi'\left(k\right)
&=
\,{1 \over 2\pi}{\cal P}\,\left[%
{32k \over k^{2}  1}

{16 \over k}\,{1 \over k^{2}  1}\,{\pi k/2 \over \tan\left(\pi k/2\right)}
\right]
\end{align}
We can observe that $\phi'\left(k\right)$ diverges at $k$ values:
$$
k = 0, \pm 1, \pm 2, \pm 4, \pm 6,\ldots
$$