If $1\leq p<2$ then $\mathscr{F}: L^p \to L^{p'}$  is not surjective.  I had this as a homework problem a week back.

The reason is the bounded inverse theorem:  $\mathscr{F}: L^p \to L^{p'}$ is injective, (by fourier inversion on the dense subspace of schwarz functions).  If the map were surjective then there would be an inverse that would be continuous, since $\mathscr{F}$ is an open map under this assumption.  

Thus we just need to prove that there is no bounded inverse:  For $f \in \mathscr{S}$, there is no constant $c$ such that $ ||f||_{p} \leq c ||\hat f||_{p'}$ for $f \in \mathscr{S}$ with the constant only depending on $p$.    

This is easy:  The function $f_\lambda=e^{-\pi i \lambda x^2-\pi x^2}$ satisfies $||f_\lambda||_p=c$ independent of $\lambda$, whereas $||\hat f_\lambda||_{p'} \leq c \lambda^{1/p'-1/2}$.  But there is no constant such that $c \leq \lambda^{1/p'-1/2} $ for all $\lambda >0$.  Therefore the fourier transform is not surjective from $L^p \to L^{p'}$ for $1\leq p<2$