I believe the argument Galatius had in mind is the following. Let us write $\bar{V} = \oplus_{n \geq 2} V_n$, so $V = V_1 \oplus \bar{V}$. All cohomology will be rational, and we write $G := \pi_1(X)$. Lets go ahead and suppose $G$ is finitely-generated and $X$ has rational cohomology of finite type.

By assumption $V_1 \to H^1(X) = H^1(G)$ is an isomorphism. As $G$ is an abelian group the extension to a free commutative graded algebra $\mathbb{Q}[V_1] \to H^*(G)$ is an isomorphism. We also have the composition $\mathbb{Q}[\bar{V}] \hookrightarrow H^*(X) \to H^*(\widetilde{X})$, and I claim this is an isomorphism in degrees $\leq n$.

To see this, we apply the spectral sequence described by Emerton in his answer, which goes
$$E_2^{p,q} = H^p(G ; H^q(\widetilde{X})) \Longrightarrow H^{p+q}(X).$$
As the action of $G$ on $H^q(\widetilde{X})$ is trivial, we can re-write $E_2$ as $H^*(G) \otimes H^*(\widetilde{X}) = \mathbb{Q}[V_1] \otimes H^*(\widetilde{X})$. The image of $\mathbb{Q}[\bar{V}] \hookrightarrow H^*(X) \to H^*(\widetilde{X})$ can support no differentials (as these classes are pulled back from $X$, by construction). The map $\mathbb{Q}[\bar{V}] \hookrightarrow H^*(X) \to H^*(\widetilde{X})$ is an isomorphism in degree 0 and 1 (this is obvious, as there is nothing in degree 1 on both sides). In the minimal degree in which it is not as isomorphism (below $n$), we see we get a contradiction. If it is not surjective, there are extra classes in $H^*(\widetilde{X})$. These must support a differential, but we know everything in lower degrees is correct, so there is nothing for such a differential to hit. Otherwise, the map is surjective but not injective. But then we do not find enough cohomology in this degree to produce the result on $E_\infty$, which we know to be $\mathbb{Q}[V]$.

So, we have that $\mathbb{Q}[\bar{V}] \to H^*(\widetilde{X})$ is an isomorphism in degrees $\leq n$. Now we use the path fibration
$$\Omega_\bullet X \simeq \Omega \widetilde{X} \longrightarrow P\widetilde{X} \longrightarrow \widetilde{X},$$
where the fibre is identified with the basepoint component of the loop space of $X$. We are now in the situation of (1) in Goodwillie's answer. We have the map $c: \mathbb{Q}[s^{-1} \bar{V}] \to H^*(\Omega_\bullet X)$ and want to see that it is an isomorphism. In low degrees the first differential is
$$d_2 : H^1(\Omega_\bullet X) \longrightarrow H^2(\widetilde{X}) = V_2$$
and this must be an isomorphism as the spectral sequence converges to zero. This shows that $c$ is an isomorphism in degree 1. Now we argue in a similar way as above by thinking about the minimal degree in which it is not an isomorphism: this argument takes us up to degree $n-1$.