How does iodide replace bromide when the latter is more basic?
As you suggest "$\ce{Br−}$  is a stronger base than $\ce{I−}$ and keeping in mind, stronger the base, lesser is the leavability, $\ce{I−}$ should not replace $\ce{Br−}$ " 

I actually agree with the logic and according to me, infact, the reverse reaction, i.e. 

$$\ce{CH3CH2CH2I + Br- (presence of DMSO) -&gt; CH3CH2CH2Br + I-}$$

should be much more favourable. Because we already know, that in the aprotic solvents, the nucleophilicity of halides is  $\ce{F− &gt; Cl− &gt; Br− &gt; I−}$, so $\ce{I−}$ should be a better leaving group and $\ce{Br−}$ should be a better nucleophile .