It seems to me that the stronger statement is true. If $f$ is only an isomorphism on homotopy in degrees $* \leq n$ then the homotopy fibre $F$ is $(n-1)$-connected and its Hurewicz ma is an isomorhism in degree $n$. Considering the Serre spectral sequence of $F \to \widetilde{X} \to \widetilde{Y}$, and using the fact that it is empty above the $n$th column, by the dimension restriction on $Y$, it follows that there is a short exact sequence

$$H_n(F) \to H_n(\widetilde{X}) \to H_n(\widetilde{Y})$$

And so $\pi_n(F) \to H_n(F) \to H_n(\widetilde{X})$ is injective. But this factors through the map $\pi_n(F) \to \pi_n(\widetilde{X})$, which is trivial.

It seems to me that the stronger statement is true. If $f$ is only an isomorphism on homotopy in degrees $* \leq n$ then the homotopy fibre $F$ is $(n-1)$-connected and its Hurewicz map is an isomorphism in degree $n$. Considering the Serre spectral sequence of the fibration seqeuence $F \to \widetilde{X} \to \widetilde{Y}$, and using the fact that it vanishes above the $n$th column (by the dimension restriction on $Y$) it follows that there is a short exact sequence $$0 \to H_n(F) \to H_n(\widetilde{X}) \to H_n(\widetilde{Y}) \to 0$$ and so the composition $\pi_n(F) \to H_n(F) \to H_n(\widetilde{X})$ is injective. But this factors through the map $\pi_n(F) \to \pi_n(\widetilde{X})$, which is trivial, and hence $\pi_n(F)=0$ too.