Why doesn't the bromide anion attack during electrophilic addition of bromine in methanol solvent?
There's a very simple explanation for this. Your reasoning regarding $\ce{Br-}$ being a better nucleophile is absolutely correct. The reason as to why methanol attacks instead of $\ce{Br-}$ is because it is the $\mathsf{\color{red}{solvent}}$. The solvent concentration is simply too high. Only one $\ce{Br-}$ is generated.   

More or less the answer is this paragraph from J. Clayden - section "Unsymmetrical bromonium ions open regioselectively" in chapter 20 "Electrophilic addition to alkenes" :


  ...when a bromination is done in a nucleophilic solvent—water or methanol, for example—solvent molecules compete with the bromide to open the bromonium ion. As
  you know, alcohols are much worse nucleophiles than bromide but, because the concentration of solvent is so high (remember—the concentration of water in water is $\pu{55M}$), the solvent gets there first most of the time. This is what happens when isobutene is treated with bromine in methanol. An ether is formed by attack of methanol only at the more substituted end of the bromonium ion.
  
  




Note: Concentration of pure methanol is around $\pu{24.75 M}$, calculated from the density given on ChemSpider