CHARGE BINDING AT THE SECOND COORDINATION SPHERE
In ["Fe"("H"_2"O")_5"NO"]"SO"_4[Fe(H2O)5NO]SO4, recall that the charge on sulfate is 2-2−. That is a good starting point. It implies the charge on ["Fe"("H"_2"O")_5"NO"][Fe(H2O)5NO] is 2+2+ to balance it out.
LIGAND CHARGES IN THE FIRST COORDINATION SPHERE
We know that water is a neutral, weak-field ligand. Thus it does not contribute to the overall charge.
The tricky part is that "NO"NO could either bind as "NO"^(+)NO+ (linear, triple-bonded) or "NO"^(-)NO− (bent, double-bonded)... and we don't really know which. This is known as a "non-innocent ligand", because it messes with us when we are trying to find the oxidation state of the transition metal.
Suppose that it was the linear "NO"^(+)NO+. Then the oxidation state on iron would be +1+1. If it was the bent "NO"^(-)NO−, then the oxidation state on iron would be +3+3. Unfortunately, both answers are available.
NARROWING DOWN WHICH OXIDATION STATE IS MORE LIKELY
Consider the electron configuration of iron, then. The atomic number is 2626, so its configuration is [Ar]3d^6 4s^2[Ar]3d64s2.
Since the sixth dd electron is paired, it is sensible for iron to have a color(blue)(+3)+3 oxidation state by losing two 4s4s and one 3d3d electron. This gives a color(blue)([Ar]3d^5)[Ar]3d5 configuration and a total electron spin of 5*"1/2" = color(blue)("5/2")5⋅1/2=5/2.
Although, it could lose one 4s4s electron and still be favorable according to Hund's rule for maximizing total electron spin when possible, giving it a color(blue)([Ar]3d^6 4s^1)[Ar]3d64s1 configuration, a +1+1 oxidation state, and a total electron spin of 5*"1/2" + ("1/2" + (-"1/2")) = color(blue)("5/2")5⋅1/2+(1/2+(−1/2))=5/2.
That is indeed still the same total electron spin, but the 3d3d orbitals are higher in energy than the 4s4s orbitals for iron, AND there is one set of electrons paired, increasing the "pairing energy" of this configuration relative to the +3+3 oxidation state (the energy associated with pairing two electrons is repulsive and thus positive).
If this configuration could be stabilized, then it would be more likely...
STABILIZING THE NORMALLY LESS-LIKELY CONFIGURATION?
If the nitrosyl ligand is "NO"^(+)NO+, then it acts as a good \mathbf(pi)-acceptor ligand (isoelectronic with "CO"), accepting electron density into its antibonding pi orbitals from a d_(xz) or d_(yz) orbital on iron via a phenomenon called "pi-backbonding".
That would indeed stabilize the d orbitals by decreasing the amount of electron density that needs to get spread around, thus decreasing the "pairing energy" (making pairing more favorable).
On the other hand, "NO"^(-), which is bent, is basically not as suitable a pi-acceptor; technically, its antibonding pi orbital is not the right "symmetry" to accept electrons from iron's d_(xz) or d_(yz) orbitals.
So, I would expect the nitrosyl ligand to be color(blue)("NO"^(+)) and for iron to be the (normally less-likely) color(blue)(+1) oxidation state.
