Can nucleophilic attack be faster than rearrangement?
You are asking what would happen if Something like 1-bromo-2-methylpropane were to undergo an $\mathrm{S_N1}$ reaction. The simple answer is: Your premise won’t happen.

The first step in an $\mathrm{S_N1}$ reaction is the formation of a carbocation by the leaving group leaving. A general equation for this would be:

$$\ce{R3C-Br &lt;=&gt; R3C+ + Br-}$$

Where this reaction has its equilibrium and thus what concentration of the carbocation can be expected, depends on the relative stabilities of reactant and carbocation (the bromide is always the same so it doesn’t matter). Now primary carbocations are so unstable (in a thermodynamic sense) that thermodynamics predicts an equilibrium far to the left for primary haloalkanes:

$$\ce{RCH2-Br &lt;&lt;=&gt; RCH2+ + Br-}$$

(This reaction arrow doesn’t do it justice. It should be a humungous left-pointing arrow and a minute right-pointing one.)

For tertiary or otherwise stabilised carbocations, we can expect a significant concentration of the right-hand side of the equation at equilibrium. Thus, we have a potential carbocation concentration that can potentially react with another nucleophile in an $\mathrm{S_N1}$ reaction. But primary haloalkanes will not even allow the carbocation to form, or in a different view, will pull back the bromide to regenerate the haloalkane before any nucleophile has a chance of attacking the carbocation.

That’s why primary haloalkanes predominantly react via the $\mathrm{S_N2}$ mechanism: There simply isn’t any significant amount of carbocation present for $\mathrm{S_N1}$ or any type of Wagner-Meerwein rearrangement.