Real and imaginary parts of $y$ form uncorrelated normal random vector with zero means, marginal variances $\frac{1}{2} \vert a \vert^2$.
Indeed, $ \mathbb{E}( \text{re}( \sum_k a_k x_k )^2 ) = \mathbb{E}( \text{im}( \sum_k a_k x_k )^2 ) = \frac{1}{2} \sum_k \vert a_k \vert^2 = \frac{1}{2} \vert a \vert^2$ and $ \mathbb{E}( \text{re}(\sum_k a_k x_k ) \text{im}( \sum_m a_m x_m ) ) = \sum_{k} \mathbb{E} (\text{re}(a_k x_k ) \text{im}( a_k x_k ) ) = \sum_k ( \text{re}(a_k) \text{im}( a_k ) - \text{im}(a_k) \text{re}( a_k ) ) = 0$.
Then your random variate $\vert y \vert$ is the some squares of two normals with variances $\vert a\vert^2$ and is thus an exponential distribution with rate $\lambda = \vert a \vert^2$, which is the same as rescaled $\chi^2_2$ variable.
Here is a simulation evidence using Mathematica: 
Here is the Mathematica code:
a = RandomComplex[1 + I, 5]; PDF[NormalDistribution[0, 1/Sqrt[2]], x] PDF[ NormalDistribution[0, 1/Sqrt[2]], y] Show[ Histogram[ Table[Abs[ a.RandomReal[NormalDistribution[0, 1/Sqrt[2]], {5, 2}].{1, I}]^2, {10^4}], Automatic, "PDF"], Plot[PDF[ExponentialDistribution[1/Norm[a]^2], x], {x, 0, 25}, PlotStyle -> Red], PlotRange -> All]