Consider all possible ways to divide $230$ labeled objects among $4$ labeled people, and for $i=0$, $\ldots$, $3$ and $j=0$, $\ldots$, $14$, let $A_{ij}$ be the set of ways where person $i$ receives exactly $j$ objects. We wish to count the size of $S$, where $S$ is the complement of $\bigcup_{i,j} A_{ij}$. Observe that $A_{ij}\cap A_{ik}=\emptyset$ if $j\ne k$. This implies that any intersection of five or more distinct $A_{ij}$'s must be empty; in fact, any intersection of four distinct $A_{ij}$'s must also be empty, since apart from those with repeated $i$'s, any such intersection will be of the form A_{0j}\cap A_{1j'} \cap A_{2j''} \cap A_{3j'''}, which is empty since it requires each person to get at most $14$ objects, but $4\cdot 14<230$. Therefore, applying the inclusion-exclusion principle will give \#S=4^{230}-\sum_{i,j} \#A_{ij}+\sum_{i,j,i',j': i -\sum_{i,j,i',j',i'',j'': i However, $\#A_{ij}$ is the number of ways to pick $j$ labeled objects out of $230$ to give to person $i$ and divide the remainder among $3$ people, so $\# A_{ij}=\binom{230}{j} 3^{230-j}.$ Similarly, \# (A_{ij}\cap A_{i'j'})=\binom{230}{j\ \ j'\ \ 230-(j+j')}2^{230-(j+j')} and \# (A_{ij}\cap A_{i'j'}\cap A_{i''j''})= \binom{230}{j\ j'\ j''\ 230-(j+j'+j'')} , so \#S =4^{230}-4 \sum_{0\le j\le 14} \binom{230}{j} 3^{230-j} +\binom{4}{2} \sum_{0\le j, j'\le 14} \binom{230}{j\ j'\ 230-(j+j')} 2^{230-(j+j')} -\binom{4}{3} \sum_{0\le j, j', j''\le 14} \binom{230}{j\ j'\ j''\ 230-(j+j'+j'')}, which can be computed to be
2977131414714304228375163768128492513800825122727620839102462174397915143246224081981112746784016599181459879137390109633489003108916312960.
As Henry says, this is very close to $4^{230}$: $1-\frac{\#S}{4^{230}}\approx 1.6848\cdot 10^{-13}.$