Your $Q$ still is four independent quantities - one for each independent choice of the translation direction $a^\nu$.
Noether's theorem states that for a one-parameter continuous quasi-symmetry \begin{align} x^\mu & \mapsto x^\mu + \epsilon \delta x^\mu \\ \phi^a & \mapsto \phi^a + \epsilon \delta \phi^a \end{align} where the theory is quasi-symmetric under this transformation in a neighbourhood of $\epsilon = 0$ for constant $\delta x^\mu, \delta \phi^a$, you get the conserved current expression in your question. The transformation \begin{align} x^\mu & \mapsto x^\mu + a^\nu \\ \phi^a & \mapsto \phi^a \end{align} is not of this form, it is really a family of one-parameter symmetries, additionally parametrized by the four-vector $a^\mu$ - it is a generic infinitesimal translation, and the translation group is $\mathbb{R}^4$. There are four independent one-parameter continuous transformations here: \begin{align} x^\mu \mapsto x^\mu + \epsilon e^0 \\ x^\mu \mapsto x^\mu + \epsilon e^1 \\ x^\mu \mapsto x^\mu + \epsilon e^2 \\ x^\mu \mapsto x^\mu + \epsilon e^3 \end{align} for $e^\mu$ the unit vector in the $\mu$-direction, and to each of these transformations Noether's theorem associates a conserved quantity. The usual derivation you cite where the $a^\nu$ is kept generic is just an efficient way to derive the conserved quantities for each of these transformations at once, where we can then choose e.g. $a = e^2$ at the end if we want the conserved quantity associated to the translation in the 2-direction.