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Warning: the following answer is (a) maybe a bit careless in a nineteenth-century sort of way, (b) missing its final step (which may be obvious to others, and/or amount to looking things up in the right table)

We are looking for an even function $g:\mathbb{R}\to \mathbb{R}$ with support on $[-1,1]$; we will define $f$ to be its Fourier transform. The condition $\int_{\mathbb{R}} f(x) dx=1$ becomes $g(0)=1$. What we need to minimize is the quantity $$I = \int_{\mathbb{R}} |x| \widehat{g}(x) dx.$$ Here are two attempts to express $I$ more directly in terms of $g$.

1. The Fourier transform of $\frac{g'(t)}{2\pi i}$ equals $x \widehat{g}(x)$. We can write $|x| \widehat{g}(x) = \mathrm{sgn}(x) x \widehat{g}(x)$. The Fourier transform of $\mathrm{sgn}(x)$ is $\frac{1}{i \pi t}$ (in some sense). Hence, the Fourier transform of $|x| \widehat{g}(x)$ should be the convolution of $\frac{1}{i \pi t}$ and $\frac{g'(-t)}{2\pi i} = \frac{g'(t)}{2\pi i}$. In particular, $I$ should equal the value of the Fourier transform of $|x| \widehat{g}(x)$ at $0$, i.e., $$I = \int_{\mathbb{R}} \frac{1}{i \pi t} \frac{g'(t)}{2\pi i} dt = - \frac{1}{2 \pi^2} \int_{\mathbb{R}} \frac{g'(t)}{t} dt.$$ We can assume $g'(0)=0$, so the integral above should make sense.

2. Since $I = - 2 \int_{-\infty}^0 x \widehat{g}(x) dx$, we can write $I = - 2\int_{-\infty}^0 \int_{-\infty}^y \widehat{g}(x) dx dy$. Now, the second antiderivative of $\widehat{g}(x)$ should have Fourier transform $\frac{g(t)}{(2\pi i)^2 t^2}$. Its value at $0$ equals both $-I/2$ (by definition) and $$\int_\mathbb{R} \frac{g(t)}{(2\pi i)^2 t^2} dt = -\frac{1}{4\pi^2} \int_{\mathbb{R}} \frac{g(t)}{t^2} dt;$$ hence, $$I = \frac{1}{2\pi^2} \int_{\mathbb{R}} \frac{g(t)}{t^2} dt.$$ Of course this diverges.

At the same time, by integration by parts, $-\int_\mathbb{R} \frac{g'(t)}{t} dt$ equals $\int_\mathbb{R} \frac{g(t)-1}{t^2} dt$, which converges. So, it looks like $$I =\frac{1}{2\pi^2} \int_\mathbb{R} \frac{g(t)-1}{t^2} dt.$$

We recall we must also fulfill the constraint that $f$ take non-negative values. This will certainly be true if $g$ is defined as a convolution $h\ast h$, with $h$ symmetric and real-valued, as then $\widehat{h}$ will be real valued, and $\widehat{h\ast h} = \widehat{h}^2$. We can require that $h$ also be an even function, and that the support of $h$ be contained in $[-1/2,1/2]$. I think these are all necessary conditions, so I am not making the search space for my optimum any smaller, but I'd be delighted if others can double-check and confirm.

So, we've reduced our problem to: find a symmetric function $h:[-1/2,1/2]\to \mathbb{R}$ with $(h\ast h)(0) = |h|_2^2 = 1$ such that $$\int_{\mathbb{R}} \frac{(h\ast h)(t)}{t^2} dt$$ is minimal.

A bit of calculus variations seems to show that the optimal $h(t)$ has to have $$\frac{1}{h(t_0)} \int \frac{\frac{1}{2} (h(t+t_0)+h(t-t_0)) - h(t_0)}{t^2} dt$$ equal to a constant independent of $t_0$, for $t_0\in (-1/2,1/2)$. Again by integration by parts, this is just $$\frac{1}{h(t_0)} \int \frac{\frac{1}{2} (h'(t+t_0)+h'(t-t_0))}{t} dt,$$ which equals $\frac{1}{h(t_0)} H(h')(t_0)$, where $H$ is our old friend the Hilbert transform.

In other words: we are to find a (continuous) function $h:\mathbb{R}\to \mathbb{R}$, supported on $[-1/2,1/2]$, with $|h|_2=1$, such that $H(h')(t) = \lambda h(t)$ for all $t\in (-1/2,1/2)$ and some $\lambda$.

Surely such a function must be known (if it exists)?

Warning: the following answer is (a) maybe a bit careless in a nineteenth-century sort of way, (b) missing its final step (which may be obvious to others, and/or amount to looking things up in the right table)

We are looking for an even function $g:\mathbb{R}\to \mathbb{R}$ with support on $[-1,1]$; we will define $f$ to be its Fourier transform. The condition $\int_{\mathbb{R}} f(x) dx=1$ becomes $g(0)=1$. What we need to minimize is the quantity $$I = \int_{\mathbb{R}} |x| \widehat{g}(x) dx.$$ Here are two attempts to express $I$ more directly in terms of $g$.

1. The Fourier transform of $\frac{g'(t)}{2\pi i}$ equals $x \widehat{g}(x)$. We can write $|x| \widehat{g}(x) = \mathrm{sgn}(x) x \widehat{g}(x)$. The Fourier transform of $\mathrm{sgn}(x)$ is $\frac{1}{i \pi t}$ (in some sense). Hence, the Fourier transform of $|x| \widehat{g}(x)$ should be the convolution of $\frac{1}{i \pi t}$ and $\frac{g'(-t)}{2\pi i} = \frac{g'(t)}{2\pi i}$. In particular, $I$ should equal the value of the Fourier transform of $|x| \widehat{g}(x)$ at $0$, i.e., $$I = \int_{\mathbb{R}} \frac{1}{i \pi t} \frac{g'(t)}{2\pi i} dt = - \frac{1}{2 \pi^2} \int_{\mathbb{R}} \frac{g'(t)}{t} dt.$$ We can assume $g'(0)=0$, so the integral above should make sense.

2. Since $I = - 2 \int_{-\infty}^0 x \widehat{g}(x) dx$, we can write $I = - 2\int_{-\infty}^0 \int_{-\infty}^y \widehat{g}(x) dx dy$. Now, the second antiderivative of $\widehat{g}(x)$ should have Fourier transform $\frac{g(t)}{(2\pi i)^2 t^2}$. Its value at $0$ equals both $-I/2$ (by definition) and $$\int_\mathbb{R} \frac{g(t)}{(2\pi i)^2 t^2} dt = -\frac{1}{4\pi^2} \int_{\mathbb{R}} \frac{g(t)}{t^2} dt;$$ hence, $$I = \frac{1}{2\pi^2} \int_{\mathbb{R}} \frac{g(t)}{t^2} dt.$$ Of course this diverges.

At the same time, by integration by parts, $-\int_\mathbb{R} \frac{g'(t)}{t} dt$ equals $\int_\mathbb{R} \frac{g(t)-1}{t^2} dt$, which converges. So, it looks like $$I =\frac{1}{2\pi^2} \int_\mathbb{R} \frac{g(t)-1}{t^2} dt.$$

We recall we must also fulfill the constraint that $f$ take non-negative values. This will certainly be true if $g$ is defined as a convolution $h\ast h$, with $h$ symmetric and real-valued, as then $\widehat{h}$ will be real valued, and $\widehat{h\ast h} = \widehat{h}^2$. We can require that $h$ also be an even function, and that the support of $h$ be contained in $[-1/2,1/2]$. I think these are all necessary conditions, so I am not making the search space for my optimum any smaller, but I'd be delighted if others can double-check and confirm.

So, we've reduced our problem to: find a symmetric function $h:[-1/2,1/2]\to \mathbb{R}$ with $(h\ast h)(0) = |h|_2^2 = 1$ such that $$\int_{\mathbb{R}} \frac{(h\ast h)(t)-1}{t^2} dt$$ is minimal.

A bit of calculus variations seems to show that the optimal $h(t)$ has to have $$\frac{1}{h(t_0)} \int \frac{\frac{1}{2} (h(t+t_0)+h(t-t_0)) - h(t_0)}{t^2} dt$$ equal to a constant independent of $t_0$, for $t_0\in (-1/2,1/2)$. Again by integration by parts, this is just $$\frac{1}{h(t_0)} \int \frac{\frac{1}{2} (h'(t+t_0)+h'(t-t_0))}{t} dt,$$ which equals $\frac{1}{h(t_0)} H(h')(t_0)$, where $H$ is our old friend the Hilbert transform.

In other words: we are to find a (continuous) function $h:\mathbb{R}\to \mathbb{R}$, supported on $[-1/2,1/2]$, with $|h|_2=1$, such that $H(h')(t) = \lambda h(t)$ for all $t\in (-1/2,1/2)$ and some $\lambda$.

Surely such a function must be known (if it exists)?

What follows is not answer, but rather a late-night, recklessly nineteenth-century exploration of a possible angle of attack. I'm pretty sure it can be improved.

We are looking for an even function $g:\mathbb{R}\to \mathbb{R}$ with support on $[-1,1]$; we will define $f$ to be its Fourier transform. The condition $\int_{\mathbb{R}} f(x) dx=1$ becomes $g(0)=1$. What we need to minimize is the quantity $$I = \int_{\mathbb{R}} |x| \widehat{g}(x) dx.$$ Here are two attempts to express $I$ more directly in terms of $g$.

1. The Fourier transform of $\frac{g'(t)}{2\pi i}$ equals $x \widehat{g}(x)$. We can write $|x| \widehat{g}(x) = \mathrm{sgn}(x) x \widehat{g}(x)$. The Fourier transform of $\mathrm{sgn}(x)$ is $\frac{1}{i \pi t}$ (in some sense). Hence, the Fourier transform of $|x| \widehat{g}(x)$ should be the convolution of $\frac{1}{i \pi t}$ and $\frac{g'(-t)}{2\pi i} = \frac{g'(t)}{2\pi i}$. In particular, $I$ should equal the value of the Fourier transform of $|x| \widehat{g}(x)$ at $0$, i.e., $$I = \int_{\mathbb{R}} \frac{1}{i \pi t} \frac{g'(t)}{2\pi i} dt = - \frac{1}{2 \pi^2} \int_{\mathbb{R}} \frac{g'(t)}{t} dt.$$ We can assume $g'(0)=0$, so the integral above should make sense.

2. Since $I = - 2 \int_{-\infty}^0 x \widehat{g}(x) dx$, we can write $I = - 2\int_{-\infty}^0 \int_{-\infty}^y \widehat{g}(x) dx dy$. Now, the second antiderivative of $\widehat{g}(x)$ should have Fourier transform $\frac{g(t)}{(2\pi i)^2 t^2}$. Its value at $0$ equals both $-I/2$ (by definition) and $$\int_\mathbb{R} \frac{g(t)}{(2\pi i)^2 t^2} dt = -\frac{1}{4\pi^2} \int_{\mathbb{R}} \frac{g(t)}{t^2} dt;$$ hence, $$I = \frac{1}{2\pi^2} \int_{\mathbb{R}} \frac{g(t)}{t^2} dt.$$

That's of course the same answer in both cases, by integration by parts, with the second integral presumably being interpreted in... som esoteric sense? (EDIT: presumably this should be $\int \frac{g(t)-1}{t^2} dt$.) Let us work with the first integral, i.e., $$I = -\frac{2}{\pi^2} \int_{\mathbb{R}} \frac{g'(t)}{t} dt.$$

We recall we must also fulfill the constraint that $f$ take non-negative values. This will certainly be true if $g$ is defined as a convolution $h\ast h$, with $h$ symmetric and real-valued, as then $\widehat{h}$ will be real valued, and $\widehat{h\ast h} = \widehat{h}^2$. We can require that $h$ also be an even function, and that the support of $h$ be contained in $[-1/2,1/2]$ (note we are adding a constraint, and so I'm concerned we are no longer finding the optimal solution to the original problem, if we decide to go this way).

At any rate, it looks like we are trying to find a symmetric function $h:[-1/2,1/2]\to \mathbb{R}$ with $(h\ast h)(0) = |h|_2^2 = 1$ such that $$\int_{\mathbb{R}} \frac{(h\ast h)(t)}{t^2} dt$$ is minimal.

Where does one take it from here? This doesn't look hard.

Warning: the following answer is (a) maybe a bit careless in a nineteenth-century sort of way, (b) missing its final step (which may be obvious to others, and/or amount to looking things up in the right table)

We are looking for an even function $g:\mathbb{R}\to \mathbb{R}$ with support on $[-1,1]$; we will define $f$ to be its Fourier transform. The condition $\int_{\mathbb{R}} f(x) dx=1$ becomes $g(0)=1$. What we need to minimize is the quantity $$I = \int_{\mathbb{R}} |x| \widehat{g}(x) dx.$$ Here are two attempts to express $I$ more directly in terms of $g$.

1. The Fourier transform of $\frac{g'(t)}{2\pi i}$ equals $x \widehat{g}(x)$. We can write $|x| \widehat{g}(x) = \mathrm{sgn}(x) x \widehat{g}(x)$. The Fourier transform of $\mathrm{sgn}(x)$ is $\frac{1}{i \pi t}$ (in some sense). Hence, the Fourier transform of $|x| \widehat{g}(x)$ should be the convolution of $\frac{1}{i \pi t}$ and $\frac{g'(-t)}{2\pi i} = \frac{g'(t)}{2\pi i}$. In particular, $I$ should equal the value of the Fourier transform of $|x| \widehat{g}(x)$ at $0$, i.e., $$I = \int_{\mathbb{R}} \frac{1}{i \pi t} \frac{g'(t)}{2\pi i} dt = - \frac{1}{2 \pi^2} \int_{\mathbb{R}} \frac{g'(t)}{t} dt.$$ We can assume $g'(0)=0$, so the integral above should make sense.

2. Since $I = - 2 \int_{-\infty}^0 x \widehat{g}(x) dx$, we can write $I = - 2\int_{-\infty}^0 \int_{-\infty}^y \widehat{g}(x) dx dy$. Now, the second antiderivative of $\widehat{g}(x)$ should have Fourier transform $\frac{g(t)}{(2\pi i)^2 t^2}$. Its value at $0$ equals both $-I/2$ (by definition) and $$\int_\mathbb{R} \frac{g(t)}{(2\pi i)^2 t^2} dt = -\frac{1}{4\pi^2} \int_{\mathbb{R}} \frac{g(t)}{t^2} dt;$$ hence, $$I = \frac{1}{2\pi^2} \int_{\mathbb{R}} \frac{g(t)}{t^2} dt.$$ Of course this diverges.

At the same time, by integration by parts, $-\int_\mathbb{R} \frac{g'(t)}{t} dt$ equals $\int_\mathbb{R} \frac{g(t)-1}{t^2} dt$, which converges. So, it looks like $$I =\frac{1}{2\pi^2} \int_\mathbb{R} \frac{g(t)-1}{t^2} dt.$$

We recall we must also fulfill the constraint that $f$ take non-negative values. This will certainly be true if $g$ is defined as a convolution $h\ast h$, with $h$ symmetric and real-valued, as then $\widehat{h}$ will be real valued, and $\widehat{h\ast h} = \widehat{h}^2$. We can require that $h$ also be an even function, and that the support of $h$ be contained in $[-1/2,1/2]$. I think these are all necessary conditions, so I am not making the search space for my optimum any smaller, but I'd be delighted if others can double-check and confirm.

So, we've reduced our problem to: find a symmetric function $h:[-1/2,1/2]\to \mathbb{R}$ with $(h\ast h)(0) = |h|_2^2 = 1$ such that $$\int_{\mathbb{R}} \frac{(h\ast h)(t)}{t^2} dt$$ is minimal.

A bit of calculus variations seems to show that the optimal $h(t)$ has to have $$\frac{1}{h(t_0)} \int \frac{\frac{1}{2} (h(t+t_0)+h(t-t_0)) - h(t_0)}{t^2} dt$$ equal to a constant independent of $t_0$, for $t_0\in (-1/2,1/2)$. Again by integration by parts, this is just $$\frac{1}{h(t_0)} \int \frac{\frac{1}{2} (h'(t+t_0)+h'(t-t_0))}{t} dt,$$ which equals $\frac{1}{h(t_0)} H(h')(t_0)$, where $H$ is our old friend the Hilbert transform.

In other words: we are to find a (continuous) function $h:\mathbb{R}\to \mathbb{R}$, supported on $[-1/2,1/2]$, with $|h|_2=1$, such that $H(h')(t) = \lambda h(t)$ for all $t\in (-1/2,1/2)$ and some $\lambda$.

Surely such a function must be known (if it exists)?

What follows is not answer, but rather a late-night, recklessly nineteenth-century exploration of a possible angle of attack. I'm pretty sure it can be improved.

We are looking for an even function $g:\mathbb{R}\to \mathbb{R}$ with support on $[-1,1]$; we will define $f$ to be its Fourier transform. The condition $\int_{\mathbb{R}} f(x) dx=1$ becomes $g(0)=1$. What we need to minimize is the quantity $$I = \int_{\mathbb{R}} |x| \widehat{g}(x) dx.$$ Here are two attempts to express $I$ more directly in terms of $g$.

1. The Fourier transform of $\frac{g'(t)}{2\pi i}$ equals $x \widehat{g}(x)$. We can write $|x| \widehat{g}(x) = \mathrm{sgn}(x) x \widehat{g}(x)$. The Fourier transform of $\mathrm{sgn}(x)$ is $\frac{1}{i \pi t}$ (in some sense). Hence, the Fourier transform of $|x| \widehat{g}(x)$ should be the convolution of $\frac{1}{i \pi t}$ and $\frac{g'(-t)}{2\pi i} = \frac{g'(t)}{2\pi i}$. In particular, $I$ should equal the value of the Fourier transform of $|x| \widehat{g}(x)$ at $0$, i.e., $$I = \int_{\mathbb{R}} \frac{1}{i \pi t} \frac{g'(t)}{2\pi i} dt = - \frac{1}{2 \pi^2} \int_{\mathbb{R}} \frac{g'(t)}{t} dt.$$ We can assume $g'(0)=0$, so the integral above should make sense.

2. Since $I = - 2 \int_{-\infty}^0 x \widehat{g}(x) dx$, we can write $I = - 2\int_{-\infty}^0 \int_{-\infty}^y \widehat{g}(x) dx dy$. Now, the second antiderivative of $\widehat{g}(x)$ should have Fourier transform $\frac{g(t)}{(2\pi i)^2 t^2}$. Its value at $0$ equals both $-I/2$ (by definition) and $$\int_\mathbb{R} \frac{g(t)}{(2\pi i)^2 t^2} dt = -\frac{1}{4\pi^2} \int_{\mathbb{R}} \frac{g(t)}{t^2} dt;$$ hence, $$I = \frac{1}{2\pi^2} \int_{\mathbb{R}} \frac{g(t)}{t^2} dt.$$

That's of course the same answer in both cases, by integration by parts, with the second integral presumably being interpreted in... som esoteric sense? (EDIT: presumably this should be $\int \frac{g(t)-1}{t^2} dt$.) Let us work with the first integral, i.e., $$I = -\frac{2}{\pi^2} \int_{\mathbb{R}} \frac{g'(t)}{t} dt.$$

We recall we must also fulfill the constraint that $f$ take non-negative values. This will certainly be true if $g$ is defined as a convolution $h\ast h$, with $h$ symmetric and real-valued, as then $\widehat{h}$ will be real valued, and $\widehat{h\ast h} = \widehat{h}^2$. We can require that $h$ also be an even function, and that the support of $h$ be contained in $[-1/2,1/2]$ (note we are adding a constraint, and so I'm concerned we are no longer finding the optimal solution to the original problem, if we decide to go this way).

Where does one take it from here?

What follows is not answer, but rather a late-night, recklessly nineteenth-century exploration of a possible angle of attack. I'm pretty sure it can be improved.

We are looking for an even function $g:\mathbb{R}\to \mathbb{R}$ with support on $[-1,1]$; we will define $f$ to be its Fourier transform. The condition $\int_{\mathbb{R}} f(x) dx=1$ becomes $g(0)=1$. What we need to minimize is the quantity $$I = \int_{\mathbb{R}} |x| \widehat{g}(x) dx.$$ Here are two attempts to express $I$ more directly in terms of $g$.

1. The Fourier transform of $\frac{g'(t)}{2\pi i}$ equals $x \widehat{g}(x)$. We can write $|x| \widehat{g}(x) = \mathrm{sgn}(x) x \widehat{g}(x)$. The Fourier transform of $\mathrm{sgn}(x)$ is $\frac{1}{i \pi t}$ (in some sense). Hence, the Fourier transform of $|x| \widehat{g}(x)$ should be the convolution of $\frac{1}{i \pi t}$ and $\frac{g'(-t)}{2\pi i} = \frac{g'(t)}{2\pi i}$. In particular, $I$ should equal the value of the Fourier transform of $|x| \widehat{g}(x)$ at $0$, i.e., $$I = \int_{\mathbb{R}} \frac{1}{i \pi t} \frac{g'(t)}{2\pi i} dt = - \frac{1}{2 \pi^2} \int_{\mathbb{R}} \frac{g'(t)}{t} dt.$$ We can assume $g'(0)=0$, so the integral above should make sense.

2. Since $I = - 2 \int_{-\infty}^0 x \widehat{g}(x) dx$, we can write $I = - 2\int_{-\infty}^0 \int_{-\infty}^y \widehat{g}(x) dx dy$. Now, the second antiderivative of $\widehat{g}(x)$ should have Fourier transform $\frac{g(t)}{(2\pi i)^2 t^2}$. Its value at $0$ equals both $-I/2$ (by definition) and $$\int_\mathbb{R} \frac{g(t)}{(2\pi i)^2 t^2} dt = -\frac{1}{4\pi^2} \int_{\mathbb{R}} \frac{g(t)}{t^2} dt;$$ hence, $$I = \frac{1}{2\pi^2} \int_{\mathbb{R}} \frac{g(t)}{t^2} dt.$$

That's of course the same answer in both cases, by integration by parts, with the second integral presumably being interpreted in... som esoteric sense? (EDIT: presumably this should be $\int \frac{g(t)-1}{t^2} dt$.) Let us work with the first integral, i.e., $$I = -\frac{2}{\pi^2} \int_{\mathbb{R}} \frac{g'(t)}{t} dt.$$

We recall we must also fulfill the constraint that $f$ take non-negative values. This will certainly be true if $g$ is defined as a convolution $h\ast h$, with $h$ symmetric and real-valued, as then $\widehat{h}$ will be real valued, and $\widehat{h\ast h} = \widehat{h}^2$. We can require that $h$ also be an even function, and that the support of $h$ be contained in $[-1/2,1/2]$ (note we are adding a constraint, and so I'm concerned we are no longer finding the optimal solution to the original problem, if we decide to go this way).

At any rate, it looks like we are trying to find a symmetric function $h:[-1/2,1/2]\to \mathbb{R}$ with $(h\ast h)(0) = |h|_2^2 = 1$ such that $$\int_{\mathbb{R}} \frac{(h\ast h)(t)}{t^2} dt$$ is minimal.

Where does one take it from here? This doesn't look hard.