The interesting point is that with factor -1/4 $$ \mathcal L =-\frac 1 4 F_{\mu\nu}F^{\mu\nu}= \frac 1 2 (\mathbf E^2- \mathbf B^2)\;, $$ From this we roughly have $$\mathcal H =\frac 1 2 (\mathbf E^2+ \mathbf B^2)\;, $$ which consistence with the energy of the EM field$^{[1]}$ $$U=\frac 1 2 \int \big( \epsilon_0 \mathbf E^2 + \frac 1 {\mu_0}\mathbf B^2 \big) d\tau \;.$$

References

[1] D.J. Griffiths, Introduction to electrodynamics Fourth Edition (See back materials)

The factor is there so that once you add a source term, i.e. $J^\mu A_\mu, $ you get the correct equations of motion, namely Maxwell's equations:

$\partial_\nu F^{\mu\nu}=J^\mu.$

Furthermore, this convention produces the usual $1/2$ in front of the kinetic term of the gauge fields.

Comments to the question:

1. First it should be stressed, as OP does, that the Euler-Lagrange equations (= classical equations of motion = Maxwell's equations) are unaffected by scaling the action $S[A]$ with an overall (non-zero) constant. So classically, one may choose any overall normalization that one would like.

2. As Frederic Brünner mentions a normalization of the $J^{\mu}A_{\mu}$ source term with a normalization constant $\pm N$ goes hand in hand with a $-\frac{N}{4}$ normalization of the $F_{\mu\nu}F^{\mu\nu}$ term. Here the signature of the Minkowski metric is $(\mp,\pm,\pm,\pm)$.

3. Recall that the fundamental variables of the Lagrangian formulation are the $4$-gauge potential $A_{\mu}$. Here $A_{0}$ is a non-dynamical Lagrange multiplier. The dynamical variables of the theory are $A_1$, $A_2$, and $A_3$. The $$-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}~=~\underbrace{\frac{1}{2} \sum_{i=1}^3\dot{A}_i\dot{A}_i}_{\text{kinetic term}}+\ldots$$ is just the standard $+\frac{1}{2}$ normalization of a kinetic term in field theory. In particular note that the kinetic term is positive definite in order not to break unitarity.