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The result the OP wants easily follows from rational homotopy theory. Let $(\Lambda W, d)$ be the minimal model of $X$, i.e it's a minimal Sullivan algebra over $\mathbb Q$ with a quasi-isomorphism $(A_{PL}X,d)\to (\Lambda W, d)$. It exists since $X$ is a nilpotent (in fact, simple) space by assumption. If you don't know what $(A_{PL}X,d)$ is, just work over $\mathbb R$ and for a smooth manifold $X$ think of the algebra of exterior differential forms on $X$. Choose a basis of $V$ and map it to closed elements of $\Lambda W$ in the corresponding cohomology classes. Extend this map multiplicatively to $\phi:(\mathbb Q[V],0)\to (\Lambda W, d)$. By assumption this map is a quasi-isomorphism up to dimension $n$. Since $(\mathbb Q[V], 0)$ is also minimal this means that $\phi$ is an isomorphism up to dimension $n$. In particular $V_i\cong \pi_i(X)\otimes \mathbb Q$ for $i\le n$. This gives homotopy groups of $\Omega X$ with degree shift by 1 and the claim follows since $\Omega X$ is an H-space and thus all the differentials in its minimal model are 0. The argument by Oscar Randal-Williams above makes all of this more explicit without relying on rational homotopy theory which has this stuff baked in. Note that by the argument above the assumptions on $X$ imply that $X$ is intrinsically formal up to dimension $n$, i.e. given its rational cohomology ring its minimal model is uniquely determined up to degree $n$. That's what makes the computation of $\pi_i(X)\otimes \mathbb Q$ particularly easy here.