**The answer below has been edited in light of other answers and comments.**

There are all sorts of models of $ZFC$ in which *every* set is definable without parameters, including nonmeasurable sets; indeed a [recent paper][1] of Hamkins, Linetsky, and Reitz is devoted to such "pointwise definable" models. 

Also, as pointed out in Theo Buehler's comment to the question, there certainly exist definable subsets of reals that are $ZFC$-provably not *Borel*.

>However, the situation is completely different for measurability. The classical work of Solovay [using an inacessible] shows that there is a model of $ZFC$ in which every subset of reals in $OD(\Bbb{R})$ is Lebesgue measurable. Recall that $X$ is in $OD(\Bbb{R})$ if $X$ is definable with parameters from $Ord \cup \Bbb{R}$.

As pointed out in Demer's answer, Krivine [without an inaccessible] provided a model of $ZFC$ in which every ordinal definable subset of reals is measurable. Moreover, as shown by Harvey Friedman, [here][2], there is a model of $ZFC$ [which is a generic extension of Solovay's model] in which the following property holds:

>**(*)**  Every equivalence class of sets of reals modulo null sets that is in $OD(\Bbb{R})$
consists of Lebesgue measurable sets.


Note that **(*)** implies that no non-measurable subset of reals in definable, since if $X$ is any definable subset of reals that is not measurable, then the equivalence class $\[X\]$ of $X$ modulo null sets satisfies the following two properties:

 **(1)** $\[X\]$ definable, 

 **(2)** No member of $\[X\]$ is measurable.

>So, to sum-up, the answer to the question for Lebesgue measurability is negative, i.e., there is no formula $\phi(x)$ in the language of set theory for which $ZFC$ proves "there is a unique nonmeasurable subset of reals satisfying $\phi$".
>
>However, if $ZFC$ is strengthened to $ZFC+V=L$ then such a formula **does** exist, as pointed out in Goldstern's answer.


  [1]: https://arxiv.org/abs/1105.4597
  [2]: http://cms.math.ca/cjm/v32/cjm1980v32.0653-0656.pdf