I am just thinking about this. So, we have the symplectic form on $\mathbb{R}^{2n}:\omega_o=\sum_{i=1}^{2n}dx_i\wedge dy_i,$ where $x_1,\dots,x_n,y_1,\dots,y_n$ are linear coordinates on $\mathbb{R}^{2n}.$ We can just use this form to get a symplectic form on the manifold. Is this correct or am I making some mistake here?
Edit: In details what I mean is suppose we have $\phi:TM\rightarrow M\times\mathbb{R}^{2n}$ which gives us an diffeomorphism such that it gives us linear isomorphism on each tangent space level. Now my proposed symplectic form is the following: for $x\in M,v,w\in T_mM,\omega_x(v,w)=\omega_0(\pi_2\circ \phi(x,v),\pi_2\circ \phi(x,w))$ where $\pi_2$ is the projection onto $\mathbb{R}^{2n}$ from $M\times\mathbb{R}^{2n}$.
A counterexample is $S^{1}\times S^{3}$. It does not have a symplectic structure because $H^{2}(S^{1}\times S^{3})=0$.
But $S^{1}\times S^{3}$ is parallelizable: both $S^{1}$ and $S^{3}$ are parallelizable since they are Lie groups, and the product of parallelizable manifolds is again parallelizable.