The assertion that there is (or is not) a truth predicate is expressible in the second-order language of set theory, but assuming consistency, not by any first-order assertion.
Second-order. In the second-order case, one simply says that there is a class $T$ satisfying the Tarskian recursion $$\exists T\ (T\text{ is a truth predicate}).$$ I gave the detailed definition of what it means to say that a class $T$ is a truth predicate in my answer to your other question, to which you linked, and those properties constitute a finite conjunction of first-order properties of $T$. So to say that there is a truth predicate involves a single second-order quantifier $\exists T$.
It follows of course that the non-existence of such a predicate is also expressible, using $$\neg\exists T\ (T\text{ is a truth predicate}).$$ This is a $\Pi^1_1$ assertion in the second-order language of set theory.
The theory GBC+"there is no truth predicate" is equiconsistent with ZFC, since clearly the consistency of this theory implies the consistency of ZFC, and conversely, if there is a model of ZFC, then there is a model of GBC having only definable classes, and this model has no truth predicate. So the assertion that there is no truth has no large-cardinal consistency strength.
In contrast, the assertion that there is a truth predicate does transcend ZFC in consistency strength, since it implies Con(ZFC) and Con(Con(ZFC)) and much more, as I explain in my blog post, to which you linked.
First-order. Meanwhile, I claim that the assertion that there is (or is not) a truth predicate, if consistent, is not expressible by any first-order assertion in the language of set theory, and perhaps this is the answer to your question.
Theorem. If GBC+$\exists$ truth-predicate is consistent, then there is no first-order assertion that GBC proves to be equivalent to the existence of a truth-predicate.
Proof. Let $(M,\in^M,S)$ be a model of GBC with a truth predicate. We may assume $M$ has a definable global well-order, by going to $L^M$ if necessary. Let $M_0=(M,\in^M,\text{Def}(M))$ be the smaller model of GBC having only definable classes. Since $M$ and $M_0$ have the same first-order objects, they satisfy all the same first-order assertions. But $M$ has a truth-predicate and $M_0$ does not. So the assertion that there is a truth predicate cannot be first-order expressible. $\Box$
Lastly, let me mention that the truth predicate, when it exists, although it is not first-order definable, is nevertheless first-order implicitly definable (and hence first-order algebraic), since when it exists it is the unique class with that first-order property. (See more about this concept of implicit definability and algebraicity in my paper: Hamkins, Joel David; Leahy, Cole, Algebraicity and implicit definability in set theory, Notre Dame J. Formal Logic, vol. 57, iss. 3, pp. 431-439, 2016. doi:10.1215/00294527-3542326, ZBL06621300.)