Consider $C_0^\infty(\mathbb R)$ as a real vector space. By choice, we can take a Hamel basis $\{e_\alpha\}_\alpha$ for $C_0^\infty(\mathbb R)$ such that every $f\in \mathbb C_0^\infty(\mathbb R)$ can be written as
$$f=\displaystyle\sum_\alpha c_\alpha e_\alpha, \quad c_\alpha \in \mathbb C.$$
Let $\{U_j\}_{j\geq 1}$ be a neighborhood basis to $0\in C_0^\infty(\mathbb R)$ in topology induced by open balls with respect to the local metric
$$d(\varphi_1, \varphi_2) := \displaystyle\sum_k 2^{-k} \frac{\|\varphi_1 - \varphi_2\|_k}{1+\|\varphi_1-\varphi_2\|_k}$$
where $$\|\varphi\|_k := \displaystyle \sum_{\alpha\leq k} \sup_{x\in K} |\partial^\alpha \varphi|, \quad \|\varphi\| := d(\varphi, 0).$$
The locality of the metric should be interpreted as defining the metric for test functions with support contained in a compact set $K \subseteq \mathbb R$.
Take now $ e_1\in \{e_α\}_α$ and $t\in \mathbb R\setminus\{0\}$ such that $t e_1 \in U_1$ and set
$$ g_1:=te_1.$$
Continue to choose $g_j\in U_j$ in the same manner. Then, by construction, all $g_j$ are linear independent and $\{ g_j\}_j$ a part of a Hamel basis. Now define the map $\lambda : C_0^\infty(\mathbb R)→\mathbb C$ with $$\lambda(g_j):= j$$ and extend $\lambda$ by linearity to be a linear map. Then $\lambda$ cannot be a distribution, since we have
$$g_j \to 0$$
in the topology on $C_0^\infty(\mathbb R)$ as defined above, but
$$\lambda(g_j) \not\to 0$$ as $\lambda$ will be an unbounded operator on any neighbourhood $U$ to $0\in C_0^\infty(\mathbb R)$.
So we have that
AoC implies the existence of a linear functional on $C_0^\infty(\mathbb R)$ which is not a distribution.
But does it require choice?