Suppose I know of all the cumulants $\{\kappa_n\}$ of some probability distribution. Is there any result of the form:
Suppose the cumulants $\{\kappa_n\}$ of a probability distribution converge to zero "sufficiently quickly". Then the distribution can be closely approximated by $\mathcal{N}(\kappa_1,\kappa_2)$
It seems that if $\kappa_3=\kappa_4=\dots=0$, then the distribution in question is $\mathcal{N}(\kappa_1,\kappa_2)$.
(As an aside, it seems that the converse isn't true. For a Poisson distribution, $\kappa_n=\lambda$ for all $n$, but for large $\lambda$, the Poisson distribution can be well-approximated by the normal distribution.)
You seem to have the wrong condition when the distribution is (almost) normal.
You should think about this analogously to some of the proofs of the central limit theorem. Given that $X$ is distributed with finite mean $$\mu = \kappa_1 = \operatorname{E}(X) $$ and finite variance $$ \sigma^2 =\kappa_2 = \operatorname{Var}(X) >0 \;.$$ We should look at the distribution of the shifted and normalized variable $$ Y = \frac{X-\mu}{\sigma}.$$
Let us denote the cumulants of $Y$ by $k_n$. By some elementary properties of the cumulant, we have that $$ \kappa_n = \sigma^n k_n, \qquad n\geq 2 \;.$$
In particular, we have that $k_1=0$ and $k_2=1$. Now the distribution of $Y$ is (almost) normal if $$ k_n \ll 1, \qquad n\geq 3.$$
By linearity, the distribution of $X$ is (almost) normal if $$ \kappa_n \ll \sigma^n = \kappa_2^{n/2},\qquad n \geq3 \;. $$
Example:
For the Poisson distribution, we have $k_n = \lambda$. The condition that the distribution is almost normal reads $$ \lambda \ll \lambda^{n/2}, \qquad n\geq 3$$ which is valid for $\lambda \gg 1$.