OK, this is not an elegant solution, but at least something...

So, let $A$ be a local $k$-algebra where $k$ is a field. I don't think that $k=\mathbb C$ makes a difference (and $k$ is easier to type than $\mathbb C$).

Anyway, the finite dimensionality condition implies that $A$ is Noetherian (in fact Artinian as the title says), so in fact it is a finitely generated $k$-algebra, in other words a surjective image of $k[x_1,\dots,x_p]$ for some $p\in \mathbb N$.

Let $I=\ker \left( k[x_1,\dots,x_p]\to A \right)$, i.e., $A\simeq k[x_1,\dots,x_p]/I$.

Now let $S\subset \mathbb N^p$ be the set of $p$-tuples $(\alpha_1,\dots,\alpha_p)$ for which $\prod x_i^{\alpha_i}\in I$, but $\prod x_i^{\alpha_i-\delta_{ij}}\not\in I$ for any $j\in\{1,\dots,p\}$ (where $\delta_{ii}=1$ and $\delta_{ij}=0$ for $i\neq j$).

Let us assume that there exists a homomorphism $\phi: k[x_1,\dots,x_p]\to k[t]/(t^n)$ such that $I=\ker \phi$. (Clearly this is equivalent to the existence of a homomorphism $A\hookrightarrow k[t]/(t^n)$.)

Let $r_i\in\mathbb N$ be such that $\phi(x_i)=u_it^{r_i}$ for some unit $u_i\in k[t]/(t^n)$.
Then it is necessary that the following inequalities are satisfied:

For any $(\alpha_1,\dots,\alpha_p)\in S$ and $j\in\{1,\dots,p\}$ , we must have

$$ n\leq \sum_i \alpha_i r_i < n + r_j $$

This certainly gives a lot of restrictions. For example this shows that the ring
$$ k[x,y]/(x^my, xy^m, x^{2m}, y^{2m})$$ where $m\geq 2$
cannot be embedded into a ring of the type $k[t]/(t^n)$. (The problem is that the first two terms imply that $mr_x+r_y\geq n$ and $r_x+mr_y\geq n$ has to hold, but this implies that then $(m+1)(r_x+r_y)\geq 2n$ which implies that then either $(m+1)r_x\geq n$ or $(m+1)r_y\geq n$ but then this contradicts $(2m-1)r_x< n$ (or the same with $r_y$) as soon as $m+1\leq 2m -1$, that is as soon as $m\geq 2$.

It seems to me that it is also a sufficient condition if the above system of inequalities has an integer solution $(r_1,\dots,r_p)$. I am not saying that it is necessarily easy to check, but at least with few variables and few relations it may not be so bad. If you want a more conceptual condition, I suppose one could associate a toric variety to the set of $S$ and there might be a nice condition of that variety that gives a condition. It will probably not be easier to check, but it might be interesting. Then again, you might not want to spend too much time on this. I think I already did. :)

**EDIT:** Corrected argument following Hailong's comment.

**Remark:** This argument is secretly about valuations...