Calculation of mass from mmol/L
There are several calcium phosphates.  I'll assume you mean monocalcium posphate, $\ce{Ca(H2PO4)2}$, because the others are not nearly soluble enough in water to allow even a 1.5 mmol/L solution.  (I assume the toothbase's most abundant ingredient is water.)

The formula weight is 234.  (I won't bother with many decimals because your range of molarities, 1.5 to 18 mmol/L, is so wide.)  I got this number from 
http://en.wikipedia.org/wiki/Monocalcium_phosphate although it is easy to compute from scratch by adding the atomic weights for 1 Ca atom, 4 H atoms etc.

To calculate how much  of this compound you need given the molarity, ideally you should be looking at a known volume of paste.  Instead you have a known weight, 1g.  If you know the density of the paste, you can compute the volume by dividing 1g by the density.  Or you can just use an approximation.  I'll assume that 1g = 0.9mL (this will be a good approximation).  From now on I will assume volume = 0.9mL.

Multiplying 1.5mmol/L by 0.9mL gives you 1.35 micro moles (1.35$\times 10^{-6}$ moles).  Since 1 mole is 234 grams (formula weight), this means 316 micrograms or 0.315 mg.  The same calculation with 18mmol/L gives 3.79 mg. 

So your answer (discarding  unwarranted decimals) is: between 0.31 and 3.8 mg .

I hope if your proprietary formulation is commercially successful you will make a generous contribution to the organization behind stackexchange!