The answer depends on how $r$ compares to the length $L$ and radius $R$ of each of the magnets.
As @klippo's answer notes, for large separations ($r>>R$) the magnetic field strength should be the dipole field that scales as $1/r^3$. (See this answer for a derivation of the $1/r^3$ law without calculus.)
If the magnets are very close together ($r << R$), then just as for the electric field of a parallel plate capacitor, the magnetic field will be constant independent of $r$. (See this answer for intuitive explanation of why fields between infinite planes are constant.)
In between, when ($R<< r << L$), each pole tip acts like a magnetic charge (monopole) and the midpoint magnetic field strength should scale as $1/r^2$, as for electric charges.
The field at the midpoint can actually be calculated exactly for identical ideal magnets.
As in the answer to How strong is the magnetic field from a neodymium magnet?, the magnetic field a distance $z$ from a pole face along the symmetry axis of an ideal cylindrical magnet is
$$B=\frac{B_r}{2}\left (\frac{L+z}{\sqrt{{R}^2+{(L+z)}^2}} -\frac{z}{\sqrt{R^2+z^2}}\right )$$
where $B_r$ is its remanence of the magnet.
For two magnets oriented as you describe, separated by a distance $r=2z$, the magnetic field strength from each magnet at the midpoint will be the same, so the combined magnetic field will just be twice that of a single magnet, i.e.
$$B(z)=B_r\left (\frac{L+z}{\sqrt{{R}^2+{(L+z)}^2}} -\frac{z}{\sqrt{R^2+z^2}}\right )$$
This has the expected constant, $1/r^2$, and $1/r^3$ dependances in the ($r<<R$), ($R<< r << L$), and ($r>>L$) limits, respectively:
$$\begin{aligned}
\lim_{z \to 0} B(z)&= B_r\frac{L}{\sqrt{{R}^2+L^2}}\\
\lim_{L \to \infty, R \to 0} B(z)&= B_r \frac{R^2}{2z^2}= B_r \frac{8R^2 }{r^2}\\
\lim_{z \to \infty} B(z)&= B_r \frac{R^2 L}{z^3}= B_r \frac{8R^2 L}{r^3}
\end{aligned}$$
See here and here for the series expansions giving the last two limits.
