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 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, 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.
Friedman's book "Fine structure and class forcing" develops forcing over ZF, rather than ZFC, in chapter 2. Although chapter 1 is about fine structure, it is not used in chapter 2. Although the rest of his book is well above my level, I find Friedman's exposition of forcing quite helpful.
Best Answer
No, it is not possible. It is consistent with ZF without choice that
From this it follows that all sets of reals are Borel. Of course, the "axiom" (*) makes it impossible to do any analysis. As soon as one allows the bit of choice that it is typically used to set up classical analysis as one is used to (mostly countable choice, but DC seems needed for Radon-Nikodym), one can implement the arguments needed to show
Logicians call the sets obtained this way $\Delta^1_1$. They are in general a subcollection of the Borel sets. To show that they are all the Borel sets requires a bit of choice (One needs that $\omega_1$ is regular).
There is actually a nice result of Suslin relevant here. He proved that the Borel sets are precisely the $\Delta^1_1$ sets: These are the sets that are simultaneously the continuous image of a Borel set ($\Sigma^1_1$ sets), and the complement of such a set ($\Pi^1_1$ sets).
That there are $\Pi^1_1$ sets that are not $\Delta^1_1$ (and therefore, via a bit of choice, not Borel) is again a result of Suslin. He also showed that any $\Sigma^1_1$ set is either countable, or contains a copy of Cantor's set and therefore has the same size as the reals. His example of a $\Sigma^1_1$ not $\Delta^1_1$ set uses logic (a bit of effective descriptive set theory), and nowadays is more common to use the example of the $\Pi^1_1$ set WO mentioned by Joel, which is not $\Delta^1_1$ by what logicians call a boundedness argument.
A nice reference for some of these issues is the book Mansfield-Weitkamp, Recursive Aspects of Descriptive Set Theory, Oxford University Press, Oxford (1985).