That reminds me of a theorem that I believe should answer your question: $\mathbf R^{2n}$ is a universal symplectic manifold for reduction:

The authors show that if a manifold $Q$ is of finite type, that is, $H^k(Q,\mathbf Z)$ is finite-dimensional, then the noncanonical cotangent bundle $(T^∗Q,dθ_Q+τ^∗_QΩ)$ can be obtained by a Marsden-Weinstein reduction of $T^∗\mathbf R^n=\mathbf R^{2n}$ relative to a torus action. Here $Ω$ is a 2-form on $Q$, $τ_Q:T^∗Q→Q$ is the bundle map and $θ_Q$ is the canonical 1-form. Since any symplectic manifold $(Q,Ω)$ is a reduction of $(T^∗Q,dθ_Q+τ^∗_QΩ)$ relative to the zero section, it follows that any symplectic manifold can be obtained by a reduction of $\mathbf R^{2n}$ with the standard symplectic structure. (etc.)

That reminds me of a paper that I believe should answer your question: $\mathbf R^{2n}$ is a universal symplectic manifold for reduction (available here):

The authors show that if a manifold $Q$ is of finite type, that is, $H^k(Q,\mathbf Z)$ is finite-dimensional, then the noncanonical cotangent bundle $(T^∗Q,dθ_Q+τ^∗_QΩ)$ can be obtained by a Marsden-Weinstein reduction of $T^∗\mathbf R^n=\mathbf R^{2n}$ relative to a torus action. Here $Ω$ is a 2-form on $Q$, $τ_Q:T^∗Q→Q$ is the bundle map and $θ_Q$ is the canonical 1-form. Since any symplectic manifold $(Q,Ω)$ is a reduction of $(T^∗Q,dθ_Q+τ^∗_QΩ)$ relative to the zero section, it follows that any symplectic manifold can be obtained by a reduction of $\mathbf R^{2n}$ with the standard symplectic structure. (etc.)