Your problem is that you've assumed the density is 1 gram/mL.
Remember that a molar is defined as a mole of solvent per liter of solution, not solvent. Usually, in introductory chemistry classes, we skip over the fact that adding solute to a solution increases its density, because it makes life more complicated.
As you just found out though, sometimes you really need to account for the change in density that occurs as you add solute. This webpage claims that the density of $12~\mathrm{M}$ sulfuric acid is about $1.634~\mathrm{g/mL}$.
Using this updated calculation, we still have the original $1177~\mathrm{g}$ of sulfuric acid. However, the solution mass is now $1000~\mathrm{mL}\cdot 1.634~\mathrm{g/mL} = 1634~\mathrm{g}$. Finally, we obtain that
$$
\frac{ 1177~\mathrm{g} \, \ce{H2SO4}}{1634~\mathrm{g} \, \mathrm{solution}} = 72.03\% \; \mathrm{w/w} $$
A note about this calculation: it's possible we're begging the question, since I don't know which values on that linked page are measured and which are calculated. It's possible that the density was calculated from measured data. The important thing here isn't how the data was obtained though, it's that the density of solution changes as you add more solute, and so you have to change how you calculate things from the molarity.
