I don't see the value you quote in your question anywhere in the tutorial. In any event, Gibb's Free energies in Gaussian will always be reported in Hartree, while the redox potential (which you would need to calculate using these Gibb's Free energies) is typically given in Volts.
To show how this works, I'll explicitly convert their example from the last section of the tutorial. Reproducing the values here:
SCF(g)(II) : SCF Done: E(UB3LYP) = -510.439067386
SCF(solv)(II): SCF Done: E(UB3LYP) = -510.444248948
GibbsCorr (II): Thermal correction to Gibbs Free Energy= 0.134232
SCF(g)(III) : SCF Done: E(UB3LYP) = -510.170289413
SCF(solv)(III): SCF Done: E(UB3LYP) = -510.239713457
GibbsCorr (III): Thermal correction to Gibbs Free Energy= 0.132483
We want to solve for $\Delta G_{ox (sol)}$ (we will later convert this to $\Delta E_{ox (sol)}$). Using Hess's law and the thermodynamic cycle above, we can write this as $$\Delta G_{\text{ox (sol)}}=\Delta G_{\text{solv}}(\text{III})+\Delta G_{\text{ox (g)}}-\Delta G_{\text{solv}}(\text{II})$$
Evaluating this in Python and converting to volts at the end:
>>> hartree_to_Volt=-2625000.5/96500 #Numerator converts hartree to J/mol
>>> dG_g=(-510.170289413+0.132483)-(-510.439067386+0.134232)
>>> dG_II=(-510.444248948--510.439067386)
>>> dG_III=(-510.239713457--510.170289413)
>>> dG_III+dG_g-dG_II
0.20278649099998347
>>> (dG_III+dG_g-dG_II)*hartree_to_Volt
-5.516213888789659
Which gives a us a value of $E_\text{red}=5.52$ (we have to flip the sign again to give a reduction potential rather than an oxidation potential). I could have simplified the math slightly here, but note that just directly computing $\Delta G_{ox (sol)}$ from the two solvated calculations would have produced a different result, as these calculations used the gas-phase geometry and did not compute thermal corrections at the optimal geometry for the solvated species.