Is the neutral di-µ-oxo-bridged manganese dimer with eight ammonia ligands a reasonable structure?
The structure is extremely reasonable to the point where I would ask why any other di-oxido-bridged manganese(II) ammine complex would not show this structure if it were proposed. To describe why, I am going to be lazy and just copy my molecular orbital scheme of a general octahedral 3d-metal complex that I have used many times before.


Scheme 1: Octahedral $\ce{[ML6]}$ complex with no &#x3c0; interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.

As you can see, the four empty orbitals ($\mathrm{a_{1g}}$ and $\mathrm{t_{1u}}$) are indeed present and part of the bond-forming orbitals, but the fact that the $\mathrm{e_g}$ orbitals are also half-filled does not prevent them from participating, too. In fact, these orbitals turn into $\mathrm{e_g^*}$ orbitals in the resulting complex, formally antibonding. This, however, doesn’t matter, because there are still many more electrons in bonding orbitals ($\mathrm{t_{1u}, a_{1g}}$ and $\mathrm{e_g}$) that this small amount of lost energy is more than offset. Typically, for a high-spin $\mathrm{d^5}$ complex, the bond order would be somewhere around 5/6. Still well within the overall bonding range.

If one looks at the kinetic stability of various complexes, one will find that $\mathrm{d^3}$ and low-spin $\mathrm{d^6}$ often give very inert octahedral complexes as no antibonding orbitals (of the simplified scheme) are populated—but even a $\mathrm{d^{10}}$ or $\mathrm{d^0}$ metal will form these complexes, their structures just won’t be as long-lasting. (Zinc has been shown to equilibrate between tetra, penta and hexacoordinated complexes in aquaeous solution as I was told in my coordination chemistry lecture.)

So yes, this octahedral complex is perfectly fine.