Calculating the molar iodine concentration of a CT contrast agent

I want to calculate the molar concentration of Iodine in a CT contrast agent.

Name: Isovue 300 (Iopamidol - $\ce{C17H22I3N3O8}$)

Molecular weight: $777.09\ \mathrm{g \ mol^{-1}}$

The part that confuses me is this: "Each $\mathrm{mL}$ of Isovue-300 (Iopamidol Injection $61%$) provides $612 \ \mathrm{mg}$ iopamidol with $1 \ \mathrm{mg}$ tromethamine and $0.39\ \mathrm{mg}$ edetate calcium disodium. The solution contains approximately $0.043\ \mathrm{mg}$ ($0.002\ \mathrm{mEq}$) sodium and $300\ \mathrm{mg}$ organically bound iodine per $\mathrm{mL}$." (source: https://www.drugs.com/pro/isovue.html)

Each molecule of Isovue contain 3 $\ce{I}$ atoms. So if I calculate the molar concentration (300 mg/ml / 777.09 g/mol * 1000 = 386.1 mM), do I need to correct for this?

I am sorry for this basic question, but I cannot find the answer on the manufacturer's website or elsewhere.

Each $\mathrm{mL}$ of Isovue-300 contains more than what you list (namely, water, sodium hydroxide, and hydrochloric acid). That means your approach is close.
The wording about the 300 $\mathrm{mg}$ of organically-bound $\ce{I}$ per $\mathrm{mL}$ of solution makes sense when you consider the atomic weight of $\ce{I}$ (about 127 $\mathrm{mg}\,\cdot\mathrm{mmol}^{-1}$) and that 3 atoms of iodine comes out to 381 $\mathrm{mg}$. Thus, for that slightly dilute solution, the 300 $\mathrm{mg}$ makes sense. It's not an additional 300 - it's what's in the $\mathrm{mL}$ as-is.
So, the molarity of $\ce{I}$ is (I abbreviate iopamidol as "iop" below to save space):
$$\left({612\,\mathrm{mg\,iop}\over 1\,\mathrm{mL}}\right)\cdot \left({3\,\mathrm{mmol}\,\ce{I}\over 777.09\,\mathrm{mg\,iop}}\right)\cdot \left({1\,\mathrm{mol}\,\ce{I}\over 1000\,\mathrm{mmol}\,\ce{I}}\right) \cdot \left({1000\,\mathrm{mL}\over 1\,\mathrm{L}}\right) = 2.36\,\mathrm{M}$$