Let $f$ be a smooth integrable function as you commented. Then we can define the Fourier Transform as
$$\mathcal{F}[f](y) := \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f(x)e^{-iyx}\mathrm{d}x$$
We can also derive the inverse of Fourier Transform as:
$$\mathcal{F}^c[f](y):=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f(x)e^{+iyx}\mathrm{d}x$$
Now lets see what happen wen we apply both to a function
$$f(y) = \mathcal{F}^c[\mathcal{F}[f]](y) = \frac{1}{2\pi}\int_\mathbb{R}\left(\int_{\mathbb{R}} f(\omega)e^{-i\omega x}\mathrm{d}\omega \right)e^{+iyx}\mathrm{d}x = \int_{\mathbb{R}}f(\omega)\left(\int_{\mathbb{R}}\frac{1}{2\pi}e^{-i(\omega-y)x}\mathrm{d}x\right)\mathrm{d}\omega$$
We can see in this last equality that the function in brakets acts as a Dirac Delta. So we can relate this as de Dirac as a notation (because the Dirac Delta just have formal sense in distribution theory)
$$\delta(\omega-y) = \frac{1}{2\pi}\int_{-\infty}^{+\infty}e^{-i(\omega-y)x}\mathrm{d}x$$
This is an intuitive constructive way of seen that the Dirac is related to Fourier Transforms. Now for your question we can use the definition on how an Dirac Delta acts
$\hat{\delta}(\omega) = \frac{1}{(2\pi)^{1/2}}\int \delta(t-t_0)e^{-i\omega t}\mathrm{d}t = \frac{1}{(2\pi)^{1/2}}e^{-\omega t_0}$
Where we just used the above
$$f(y) = \int f (\omega)\delta(\omega-y)\mathrm{d}\omega$$
For $f(t):=e^{-i\omega t}/\sqrt{2\pi}$
In the language of distributions, the Dirac delta distribution is the map $\delta$ from the space of test functions (smooth compactly supported functions) to, say $\mathbb{R}$ with the "operation" $(\delta, f) = f(0)$ for every test function $f$.
To figure out the Fourier transform of a distribution, you need to determine the Fourier transform of a test function $f$.
$$\widehat{f}(\xi) = \frac{1}{\sqrt{2\pi}}\int_\mathbb{R} e^{-ix \xi}f(x) \, dx$$
By definition, the Fourier transform of a distribution $\varphi$ is defined by $(\widehat{\varphi},f)=(\varphi,\widehat{f})$ for every test function $f$.
EDIT: As commenters below pointed out, I should say Schwartz function instead of test function and tempered distribution instead of distribution..
Therefore
$$(\widehat{\delta},f) = (\delta,\widehat{f}) = \widehat{f}(0) = \frac{1}{\sqrt{2\pi}}\int_\mathbb{R} e^{-i x \cdot 0} f(x) \, dx =\frac{1}{\sqrt{2\pi}}\int_\mathbb{R} f(x) \, dx = (\frac{1}{\sqrt{2\pi}},f)$$
where the last equality is because the "constant" distribution 1 is regular, i.e., can be represented in integral form. Therefore, as a distribution, $\widehat{\delta} = (2\pi)^{-1/2}$.
Best Answer
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 ...)