Let me expand Nate Eldredge's comments. A more elementary proof than the one provided by Jochen Wengenroth, still requiring no computations, is as follows:

$\mathcal{E}'$ is the dual space of $C^\infty$, which is endowed with the Fréchet space topology given e.g. by the seminorms $\|f\|_k:=\|f\|_{C^k(B_k(0))}$. The fact that $f_n\to f$ in $\mathcal{E}'$ means that $\langle f_n,\phi\rangle\to\langle f,\phi\rangle$ for any $\phi\in C^\infty$.

Now the uniform boundedness principle holds also for linear maps from a Fréchet space to (say) a normed vector space. The proof is the same as that for the Banach space version; for details, see Theorem 2.6 in Rudin's Functional Analysis (2nd edition). Hence the $f_n$'s are equicontinuous, i.e. there exists $k\in\mathbb{N}$ and $C>0$ such that
$$ |\langle f_n,\phi\rangle|\le C\|\phi\|_{C^k(B_k(0))}. $$
This gives $|\widehat{f_n}(\xi)|\le C\|e^{-2\pi i\langle\xi,\cdot\rangle}\|_{C^k(B_k(0))}\le C'(1+|\xi|^k)e^{2\pi k|\xi|}$ for $\xi\in\mathbb{C}$. Also, $\widehat{f_n}(\xi)\to\widehat{f}(\xi)$ for every $\xi\in\mathbb{C}$ (since it corresponds to evaluation of $f_n$ on the test function $e^{-2\pi i\langle\xi,\cdot\rangle}$). So you have a locally equibounded sequence of pointwise converging holomorphic functions and this implies your claim. This clearly works in any dimension.

Functional Analysis(Th. 7.23, p. 183 in my edition). Is (x) "for free"? $\endgroup$ – TaQ Sep 19 '18 at 8:11