What is the Fourier transform of $|x|$?

From $operatorname{sign}'=2delta$ we get $ixi,widehat{operatorname{sign}}(xi)=2,$ from which we can conclude $$ widehat{operatorname{sign}}(xi) = -2ioperatorname{pv}frac{1}{xi}+Cdelta(xi). $$ Since $operatorname{sign}$ is odd, so must be $widehat{operatorname{sign}},$ which forces $C=0.$

Now, $f(x) = x operatorname{sign}(x),$ so $$ hat{f}(xi) = ifrac{d}{dxi}widehat{operatorname{sign}}(xi) = ifrac{d}{dxi}left(-2i operatorname{pv}frac{1}{xi}right) = -2 operatorname{fp}frac{1}{xi^2}. $$

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 $f(x)=|x|=lim_{ato 0}frac{1-e^{-a|x|}}{a}$, then one finds that in distribution

$$mathscr{F}{f}(omega)=lim_{ato 0^+}frac1aleft(2pidelta(omega)-frac{2a}{a^2+omega^2}right)$$

To evaluate this distributional limit, we begin with a test function $phi(omega)$ and find

$$begin{align} langle mathscr{F}{f},phirangle&=lim_{ato0^+}frac1aleft(2piphi(0)-int_{-infty}^infty frac{2aphi(omega)}{a^2+omega^2},domegaright)\\ &=lim_{ato 0^+}int_{-infty}^infty left(-frac{2(phi(omega)-phi(0))}{a^2+omega^2}right),domega\\ &=-lim_{varepsilonto 0^+}int_{|omega|ge varepsilon}frac{2(phi(omega)-phi(0))}{omega^2}tag{1E} end{align}$$

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 $mathscr{F}{f}(omega)=int_{-infty}^infty f(x)e^{iomega x},dx$. Had we used instead the convention $mathscr{F}{f}(omega)=int_{-infty}^infty f(x)e^{i2pi xi x},dx$, then $(2E)$ would be replaced with $-frac1{2pi^2 xi^2}$

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In This Answer, I showed that the Fourier Transform of $f(t)=tH(t)$, where $H(t)$ denotes the Heaviside function, is given by

$$mathscr{F}{f}(omega)=-frac1{omega^2}+ipi delta'(omega)tag1$$

where the distribution $d(omega)=displaystyle -frac1{omega^2}$ in $(1)$ is interpreted to mean

$$langle d, phirangle=-lim_{varepsilonto0^+}int_{|omega|gevarepsilon}frac{phi(omega)-phi(0)}{omega^2},domegatag2$$

where $phi(omega)$ is a Schwartz function.

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Using $g(t)=ttext{sgn}(t)=2tH(t)-t$ along with $mathscr{F}{t}(omega)=i2pi delta'(omega)$ 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!

So, a way to compute it is to write $|x| = xmathop{mathrm{sign}}(x)$. By definition, we have $$ langle mathcal{F}(|x|),varphirangle = langle |x|,mathcal{F}(varphi)rangle = langle xmathop{mathrm{sign}}(x),mathcal{F}(varphi)rangle $$ Since $x∈ C^infty$, we can then write $$ langle xmathop{mathrm{sign}}(x),mathcal{F}(varphi)rangle = langle mathop{mathrm{sign}}(x),x,mathcal{F}(varphi)rangle = frac{1}{2ipi}langle mathop{mathrm{sign}}(x),mathcal{F}(varphi')rangle $$ where I used the formula for the Fourier transform of a derivative. Now, by definition again, and then using the fact that $mathcal{F}(mathop{mathrm{sign}}(x)) = 1/{ipi} ,mathrm{P}(tfrac{1}{x})$ (the principal value of $1/x$) we get $$ frac{1}{2ipi}langle mathop{mathrm{sign}}(x),mathcal{F}(varphi')rangle = frac{1}{2ipi}langle mathcal{F}(mathop{mathrm{sign}}(x)),varphi'rangle \ = frac{-1}{2pi^2}langle mathrm{P}(tfrac{1}{x}),varphi'rangle = frac{1}{2pi^2}langle mathrm{P}(tfrac{1}{x})',varphirangle $$ so that $$ mathcal{F}(|x|) = frac{1}{2pi^2} mathrm{P}(tfrac{1}{x})' = frac{-1}{2pi^2} mathrm{P}(tfrac{1}{x^2}) $$ 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 $$ mathcal{F}(|x|) = frac{-1}{2pi^2x^2} $$ (if I did not make mistakes in the constants and signs ...)