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Corrected argument for $L^1$ norms

For the $L^\infty$ norms, nothing at all apart from Alexander Eremenko's answer, because of Borel's theorem: any sequence of real (resp. complex) numbers can be realized as the sequence of Taylor coefficients of some real (resp. complex) valued smooth function $f$ at any given point. By the Sobolev embedding theorem and the fact that $f$ may be chosen compactly supported in the interior of $\Omega$, one can bound above the $L^\infty$ norm of $f$ by a positive constant times the $L^1$ norm of the $n$-th order derivative of $f$, where $n$ is the dimension of the domain $\Omega$. Since you assumed $\Omega$ to be bounded, this implies the same property for the $L^p$ norms for all $1\leq p\leq\infty$.