No.

Even for $n=0$, take $Y = \mathbb A^m$, $X = \mathbb A^m$ minus a point, $f$ the inclusion.

Then $R^q f_* \mathbf Q_\ell$ will be nonvanishing for $q = 2m-1$.

$Y$ admits a smooth surjective morphism from a scheme $Z$. Because smooth morphisms are locally of finite type, $Z \to Y$ is locally of finite type, and you can choose an open cover that covers $Y$ and then pass to a finite subcover to make $Z$ of finite type.

Because this morphism is smooth, by smooth base change the pullback of $R^q f_* \mathbf Q_\ell$ to $Z$ is the pushforward of $\mathbb Q_\ell$ from $Z \times_Y X$ to $Z$. Because this morphism is smooth, it suffices to prove a bound for this pushforward.

If $Z \to Y$ is a schematic morphism (this might be a little stronger than the fibers being schemes) then $Z \times_Y X$ is a scheme, also of finite type. Boundedness then follows from classical results - mod $\ell$ cohomology is a limit over the cohomology of neighborhoods, and these are finite type of bounded dimension.