My initial intuition was incorrect. It seems we can sometimes solve this explicitly, and the minimizer does not have full support; my argument is incomplete (as discussed below), but I think it's reasonably convincing anyway. I'm also only going to discuss $a=1$ (initially, I thought that was the general case, but of course that's not true because of the $1$ in the denominator of the kernel; see also Mateusz's answer for more on this). Then it is given by
$$
\mu= \frac{5}{12}(\delta_{-1}+\delta_1) + \frac{1}{6}\delta_0 .
$$
To see why this is optimal, look at
$$
F(x)=\int \frac{d\mu(t)}{1+(t-x)^2} = \frac{5}{12}\left( \frac{1}{1+(x+1)^2} + \frac{1}{1+(x-1)^2} \right) + \frac{1}{6}\frac{1}{1+x^2} .
$$
Notice that $F(x)=7/12$ on $x=0,\pm 1$ (the support of $\mu$), and $F(x)\ge 7/12$ for $-1\le x\le 1$ (this is a tedious but elementary calculus exercise).

This means that if we vary the measure slightly, say $\nu=\mu+\epsilon\sigma$ with a signed measure $\sigma$ with $\int d\sigma=0$ and $\epsilon\ll 1$, then the change in first order will be proportional to $\int F(x)\, d\sigma(x)$. However, since $\nu$ must remain a positive measure, the negative part of $\sigma$ can only be supported by $0,\pm 1$. This means that $\int F\, d\sigma\ge 0$, by the properties of $F$ observed above.

This verifies that my $\mu$ is a local minimum. I don't have an argument that shows that it gives the global minimum also, though I'm fairly optimistic that this will be true.

A few things can perhaps be said in general: First of all, this criterion ($F_{\mu}$ constant on the support of $\mu$ and $F(x)$ at least as large otherwise) is also necessary for local minima. The general scenario that seems likely is that as you increase $a$, additional points enter the support. This would also be consistent with what Mateusz does in his answer.

Finally, here is a general argument why a measure $d\mu =f\, dx$ with $f>0$ on $-a<x<a$ can not give a minimum, not even a local one. To see this, consider again a small variation $d\nu =(f+\epsilon g)\, dx$. Now $g$ can be an arbitrary (let's say, almost arbitrary, to be safe) function with $\int g=0$, so it now follows that at an extremum, $\int Fg =0$ for all such $g$. This forces $F$ to be constant on $-a<x<a$, but since $F$ is the restriction of a harmonic (on $\mathbb C^+$) function to the line $\textrm{Im}\: z =1$ (your kernel is the Poisson kernel for the upper half plane), this harmonic function would have to be constant on this whole line, which makes $f$ constant on $\mathbb R$, but we need an $f$ that is zero outside $(-a,a)$.

nota potential kernel of any Lévy process (if it were a potential kernel, the generator of the corresponding Markov process would be an operator with Fourier symbol $\exp(|x|)$, which is impossible). Interestingly, even in the limiting cases $a \to \infty$ or $a \to 0^+$ there are no reasonable Markov processes behind this functional. $\endgroup$Intégrales de Riemann–Liouville et potentiels, if only to see how the theory developed almost a century ago. A completely general theory can be found in any books on general potential theory, of which I know mostly Bliedtner and Hansen'sPotential Theory An Analytic and Probabilistic Approach to Balayage(which is terrible to read, but I like the non-mathematical flavour of the title) and... $\endgroup$Markov Processes, Brownian Motion, and Time Symmetry(a fantastic read, but rather technical and written from a different perspective). $\endgroup$8more comments