In any structure, the non-definable elements are exactly the elements that realize the type $p(x)$, which is the type containing all assertions of the form: 

$$\varphi(x)\to\exists y\neq x\ \varphi(y).$$

The reason is that if $x$ is not definable in a structure $\mathcal{M}$, then it is not the unique satisfying instance of any formula (for this is what it means to be definable), and so if $\varphi(x)$ holds, then there must be some other $y$ also satisfying that formula. And conversely, if $x$ is definable, then there is some formula $\varphi$ for which $x$ is the only satisfying instance of $\varphi(x)$. 

For some theories, this type is principal, such as in the theory of an infinite set, since in all models of this theory, there are no definable elements, and so the type is generated by $x=x$.

But in your case of set theory, the type is not principal. To see this, suppose it were principal, generated by $\varphi(x)$. Take any pointwise definable model $\mathcal{M}$ of set theory, which is a model in which every object is definable without parameters (see my paper [Pointwise definable models of set theory](http://jdh.hamkins.org/pointwisedefinablemodelsofsettheory/)), and let $\mathcal{M}^+$ be any elementary extension of $\mathcal{M}$. So $\mathcal{M}^+$ satisfies $\exists x\varphi(x)$, since it must have non-definable elements, but $\mathcal{M}$ does not satisfy this assertion, since every element of $\mathcal{M}$ is definable there.