Why don't ligands affect the oxidation state of the central metal ion?
If you take a glimpse at the molecular orbital scheme, you will notice that the electrons donated by the water molecules still reside in orbitals that are largely ligand-centred (and have a relatively low energy). On the other hand, iron(III) also has five d electron itself; these reside in the $\mathrm{t_{2g}}$ and $\mathrm{e_g}^*$ orbitals, which are largely metal-centred (and have a relatively high energy).

This is true for most complexes. The metal tends to have rather high-lying orbitals while the ligands tend to donate rather low-lying ones to the overall complex. As long as this order is retained, there is no chemical redox behaviour going on.

For some systems, however, the orbitals are either so similar or even crossed-over that suddenly a metal orbital will become lower-lying than a ligand orbital. In this case, an electron that used to reside in a ligand orbital will relax into a metal orbital — a redox process. You can for example think of the reaction of iodide with copper(II) in this manner: copper(II) will immediately get reduced to copper(I) by iodide ions. While this is a unidirectional process, some systems are also known to bind ligands in a redox fashion reversively, meaning that the electrons will be exchanged both ways. The binding mode of oxygen to haemoglobin is understood in this manner.

Finally, sometimes there is a change in oxidation state that is merely formal. Tetracarbonylhydridoiron(0) $\ce{[Fe(CO)4H]}$ is such a case. It can be protonated to formally give tetracarbonyldihydridoiron(II) $\ce{[Fe(CO)4H2]+}$ or deprotonated to give tetracarbonylferrate(–II) $\ce{[Fe(CO)4]-}$. Both changes in oxidation state are merely formal and chemically all that is happening is a protonation/deprotonation.