It is the case that $n \in \{0,1\}$. (I include $n=0$ because $0$-manifolds with $\omega = 0$ are symplectic.)

Two-sentence summary: We know what integral elements of cohomology are, and pullback acts integrally. This allows us to think of the coefficients of $\omega$ as living in some fixed ring extension of $\mathbb{Z}$, and derive a number-theoretic contradiction.

Details: Consider the action of the pullback of these maps on cohomology, $$\phi_k^*~ \colon H^2(M;\mathbb{Z})/\mathrm{Tors} \rightarrow H^2(M;\mathbb{Z})/\mathrm{Tors},$$ which upon tensoring with $\mathbb{R}$ yields the pull-back map $$\phi_k^* \otimes \mathbb{R}~\colon H^2(M;\mathbb{R}) \rightarrow H^2(M;\mathbb{R}).$$ (Typically one just denotes this as $\phi_k^*$, but we make explicit that it comes from an integral map.) You are asking that $\omega$ is an eigenvector of $\phi_k^* \otimes \mathbb{R}$ with eigenvalue $\sqrt[n]{k}$.

For simplicity, pick a $\mathbb{Z}$-basis $(e_1,\ldots,e_r)$ for $H^2(M;\mathbb{Z})/\mathrm{Tors}$ (where $r = \beta_2$ is the 2nd Betti number), which therefore also determines a basis for $H^2(M;\mathbb{R})$. We may write $$\omega = \sum_{i=1}^{r} c_i e_i.$$ Notice that the elements $c_i$ define a ring $\mathbb{Z}[c_1,\ldots,c_r]$. (In fact, this ring is independent of the basis, which, to repeat, was only included for simplicity.) As the maps $\phi_k^*$ act integrally on the $e_i$, we have that the coefficients of $$(\phi_k^* \otimes \mathbb{R})(\omega) = \sqrt[n]{k} \cdot \omega = \sum_{i=1}^{r} (\sqrt[n]{k} \cdot c_i)e_i$$ also lie in the same ring, i.e. $\sqrt[n]{k} \cdot c_i \in \mathbb{Z}[c_1,\ldots,c_r]$. We hence find that $$\sqrt[n]{k} \in \mathbb{Q}(c_1,\ldots,c_r).$$ This must hold for all $k$ simultaneously, or in other words, $$\mathbb{Q}(\sqrt[n]{2},\sqrt[n]{3},\ldots) \subseteq \mathbb{Q}(c_1,\ldots,c_r).$$ So the field $\mathbb{Q}(\sqrt[n]{2},\sqrt[n]{3},\ldots)$ must be finitely generated over $\mathbb{Q}$, and this implies $n \in \{0,1\}$.

It is the case that $n \in \{0,1\}$. (I include $n=0$ because $0$-manifolds with $\omega = 0$ are symplectic.)

Two-sentence summary: We know what integral elements of cohomology are, and pullback acts integrally. This allows us to think of the coefficients of $\omega$ as living in some fixed abelian subgroup of $\mathbb{R}$ of finite rank, which yields a contradiction.

Details: Consider the action of the pullback of these maps on cohomology, $$\phi_k^*~ \colon H^2(M;\mathbb{Z})/\mathrm{Tors} \rightarrow H^2(M;\mathbb{Z})/\mathrm{Tors},$$ which upon tensoring with $\mathbb{R}$ yields the pull-back map $$\phi_k^* \otimes \mathbb{R}~\colon H^2(M;\mathbb{R}) \rightarrow H^2(M;\mathbb{R}).$$ (Typically one just denotes this as $\phi_k^*$, but we make explicit that it comes from an integral map.) You are asking that for each $k$, $\omega$ is an eigenvector of $\phi_k^* \otimes \mathbb{R}$ with eigenvalue $\sqrt[n]{k}$.

For simplicity, pick a $\mathbb{Z}$-basis $(e_1,\ldots,e_r)$ for $H^2(M;\mathbb{Z})/\mathrm{Tors}$ (where $r = \beta_2$ is the 2nd Betti number), which therefore also determines a basis for $H^2(M;\mathbb{R})$. We may write $$\omega = \sum_{i=1}^{r} c_i e_i,$$ where WLOG $c_1 \neq 0$. The elements $c_i$ generate a subgroup $G \leq \mathbb{R}$ (which is a free abelian group of finite rank at most $r$), and since the maps $\phi_k^*$ act integrally on the $e_i$, we have that the coefficients of $(\phi_k^* \otimes \mathbb{R})(\omega)$ with respect to the chosen basis lie again in $G$. But also $$(\phi_k^* \otimes \mathbb{R})(\omega) = \sqrt[n]{k} \cdot \omega = \sum_{i=1}^{r} (\sqrt[n]{k} \cdot c_i)e_i.$$ In particular, looking at the first coefficient, we have $\sqrt[n]{k} \cdot c_1 \in G$ for each $k$. Dividing by $c_1$, it follows that $\sqrt[n]{k}$ lies in the subgroup $\frac{1}{c_1}G$ for every $k$. But when $n \geq 2$, the elements $\sqrt[n]{k}$ generate an infinite rank subgroup of $\mathbb{R}$, so it is impossible for all of these elements to lie in the finite rank group $\frac{1}{c_1}G$. Thus $n \in \{0,1\}$.