What is the Fourier transform of $|x|$?
So, a way to compute it is to write $|x| = x\mathop{\mathrm{sign}}(x)$. By definition, we have $$ \langle \mathcal{F}(|x|),\varphi\rangle = \langle |x|,\mathcal{F}(\varphi)\rangle = \langle x\mathop{\mathrm{sign}}(x),\mathcal{F}(\varphi)\rangle $$ Since $x∈ C^\infty$, we can then write $$ \langle x\mathop{\mathrm{sign}}(x),\mathcal{F}(\varphi)\rangle = \langle \mathop{\mathrm{sign}}(x),x\,\mathcal{F}(\varphi)\rangle = \frac{1}{2i\pi}\langle \mathop{\mathrm{sign}}(x),\mathcal{F}(\varphi')\rangle $$ where I used the formula for the Fourier transform of a derivative. Now, by definition again, and then using the fact that $\mathcal{F}(\mathop{\mathrm{sign}}(x)) = 1/{i\pi} \,\mathrm{P}(\tfrac{1}{x})$ (the principal value of $1/x$) we get $$ \frac{1}{2i\pi}\langle \mathop{\mathrm{sign}}(x),\mathcal{F}(\varphi')\rangle = \frac{1}{2i\pi}\langle \mathcal{F}(\mathop{\mathrm{sign}}(x)),\varphi'\rangle \\ = \frac{-1}{2\pi^2}\langle \mathrm{P}(\tfrac{1}{x}),\varphi'\rangle = \frac{1}{2\pi^2}\langle \mathrm{P}(\tfrac{1}{x})',\varphi\rangle $$ so that $$ \mathcal{F}(|x|) = \frac{1}{2\pi^2} \mathrm{P}(\tfrac{1}{x})' = \frac{-1}{2\pi^2} \mathrm{P}(\tfrac{1}{x^2}) $$ where $\mathrm{P}(\tfrac{1}{x^2})$ is the Hadamard finite part of $\tfrac{1}{x^2}$. Away from $0$, we can thus say that $$ \mathcal{F}(|x|) = \frac{-1}{2\pi^2x^2} $$ (if I did not make mistakes in the constants and signs ...)
EDIT: I reread the post and wanted to present an edit to the original posted solution that addresses directly the concern about the OP's analysis. To that end we proceed with the addendum.
You were on the right track! In fact, if one begins with the regularization $f(x)=|x|=\lim_{a\to 0}\frac{1-e^{-a|x|}}{a}$, then one finds that in distribution
$$\mathscr{F}\{f\}(\omega)=\lim_{a\to 0^+}\frac1a\left(2\pi\delta(\omega)-\frac{2a}{a^2+\omega^2}\right)$$
To evaluate this distributional limit, we begin with a test function $\phi(\omega)$ and find
$$\begin{align} \langle \mathscr{F}\{f\},\phi\rangle&=\lim_{a\to0^+}\frac1a\left(2\pi\phi(0)-\int_{-\infty}^\infty \frac{2a\phi(\omega)}{a^2+\omega^2}\,d\omega\right)\\\\ &=\lim_{a\to 0^+}\int_{-\infty}^\infty \left(-\frac{2(\phi(\omega)-\phi(0))}{a^2+\omega^2}\right)\,d\omega\\\\ &=-\lim_{\varepsilon\to 0^+}\int_{|\omega|\ge \varepsilon}\frac{2(\phi(\omega)-\phi(0))}{\omega^2}\tag{1E} \end{align}$$
So, we find that
$$\mathscr{F}\{f\}(\omega)=-\frac2{\omega^2}\tag{2E}$$
where we interpret the distribution in $(2E)$ in the sense of $(1E)$
Note that we have used the convention $\mathscr{F}\{f\}(\omega)=\int_{-\infty}^\infty f(x)e^{i\omega x}\,dx$. Had we used instead the convention $\mathscr{F}\{f\}(\omega)=\int_{-\infty}^\infty f(x)e^{i2\pi \xi x}\,dx$, then $(2E)$ would be replaced with $-\frac1{2\pi^2 \xi^2}$
In This Answer, I showed that the Fourier Transform of $f(t)=tH(t)$, where $H(t)$ denotes the Heaviside function, is given by
$$\mathscr{F}\{f\}(\omega)=-\frac1{\omega^2}+i\pi \delta'(\omega)\tag1$$
where the distribution $d(\omega)=\displaystyle -\frac1{\omega^2}$ in $(1)$ is interpreted to mean
$$\langle d, \phi\rangle=-\lim_{\varepsilon\to0^+}\int_{|\omega|\ge\varepsilon}\frac{\phi(\omega)-\phi(0)}{\omega^2}\,d\omega\tag2$$
where $\phi(\omega)$ is a Schwartz function.
Using $g(t)=t\text{sgn}(t)=2tH(t)-t$ along with $\mathscr{F}\{t\}(\omega)=i2\pi \delta'(\omega)$ and $(1)$, we find that
$$\begin{align} \mathscr{F}\{g\}&=-\frac2{\omega^2}\tag3 \end{align}$$
where again $(3)$ is defined analogously to $(2)$.
And we are done!
From $\operatorname{sign}'=2\delta$ we get $i\xi\,\widehat{\operatorname{sign}}(\xi)=2,$ from which we can conclude $$ \widehat{\operatorname{sign}}(\xi) = -2i\operatorname{pv}\frac{1}{\xi}+C\delta(\xi). $$ Since $\operatorname{sign}$ is odd, so must be $\widehat{\operatorname{sign}},$ which forces $C=0.$
Now, $f(x) = x \operatorname{sign}(x),$ so $$ \hat{f}(\xi) = i\frac{d}{d\xi}\widehat{\operatorname{sign}}(\xi) = i\frac{d}{d\xi}\left(-2i \operatorname{pv}\frac{1}{\xi}\right) = -2 \operatorname{fp}\frac{1}{\xi^2}. $$