Why do polar aprotic solvents favour SN2 over E2?
Competition between substitution and elimination is one of the many points of confusion for introductory organic students.
In general, they are taught that there are four factors that will determine the outcome of a reaction where there may be competition between substitution or elimination.
They are:  


Type alkyl halide (methyl, $1^o$, $2^o$ or $3^o$)
The strength/basicity of the nucleophile
The solvent
The temperature


I think that many students over-emphasize the impact of any one of these factors.  It is more useful to look at the overall impact of the factors and evaluate the reaction as a whole.

Let's examine the case in question:


Primary alkyl halide - generally favours substitution
The nucleophile is $\ce{^-CN}$ which is a good nucleophile but a weak base.
Polar aprotic solvent
Low temp.


The big factor here is not the solvent.  All of the factors point toward substitution, especially the alkyl halide and the nucleophile.  Remember that elimination reactions are essentially acid-base in nature.  If the nucleophile is not a strong base, elimination reactions are less likely.