Let me first consider the case where we use the usual notion of projectively definable set of reals, where $\varphi(x)$ is a property of the reals that might become a member of the set $X$ we are defining. Suppose that $X=\{ x\in\mathbb{R}\mid \varphi(x,z)\}$ is a projective set of reals, defined with a projective formula $\varphi$, which means that all quantifiers range only over reals and natural numbers, in a suitable language, with real parameter $z$. (One doesn't ordinarily allow ordinal parameters in a projective definition, and it isn't clear how one could use them anyway if all the quantifiers only range over reals and natural numbers.)

To say that $X$ is measurable is equivalent to the assertion that we may cover $X$ and its complement with open sets, whose intersection has as small a measure as we like: for every positive $\epsilon$, there are open sets $U\supset X$ and $V\supset\mathbb{R}\setminus X$ with $\mu(U\triangle V)\lt\epsilon$. Quantifying over open sets amounts to a real quantifier, since every open set is a countable union of rational intervals; and we may take $\epsilon=\frac1n$.

Thus, if $X$ has complexity $\Delta^1_n$ in the projective hierarchy, then the assertion "$X$ is measurable" has complexity at most $\Sigma^1_{n+1}$, and the assertion "$X$ is not measurable" has complexity at most $\Pi^1_{n+1}$, and these assertions are only a little bit more complicated than $X$ itself. In particular, these assertions are also projective.

For your second question, you seem to be asking whether there is a particular projective definition of a set of reals that we can prove (from ZFC?) that it is definitely not measurable. The answer is that we should not expect to do this, because the existence of certain large cardinals implies that every projective sets of reals is Lebesgue measurable, and so succeeding in your task would refute the consistency of those large cardinals, which would be an unexpected and amazing development. Meanwhile, of course, we cannot say for certain that we cannot find an example of what you seek, since the nonexistence of such a formula implies at the very least that ZFC is consistent, which is also something that, if true, we cannot prove.

Now let us consider your rather more idiosyncractic notion of definablity, where $\varphi(X)$ is a property of the set $X$ of reals, perhaps defined using only quantification over natural numbers and reals, where we allow $X$ to appear as a predicate in atomic subformulas of $\varphi$. The point to make now is that "$X$ is nonmeasurable" is expressible using quantifiers only over the natural numbers and the reals. So the family of all non-measurable sets is definable by a definition of your favored type. In this case, we can say, yes, indeed, there is a formula $\varphi(X)$ which ZFC proves holds of some sets, such that ZFC proves that $\forall X\ \varphi(X)\to X$ is not measurable, since in fact ZFC proves $\varphi(X)\leftrightarrow X$ is not measurable.