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](https://mathoverflow.net/a/273121/1946), 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](http://jdh.hamkins.org/km-implies-conzfc/), 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: <cite authors="Hamkins, Joel David; Leahy, Cole">_Hamkins, Joel David; Leahy, Cole_, [**Algebraicity and implicit definability in set theory**](http://jdh.hamkins.org/algebraicity-and-implicit-definability/), Notre Dame J. Formal Logic, vol. 57, iss. 3, pp. 431-439, 2016. [doi:10.1215/00294527-3542326](http://dx.doi.org/10.1215/00294527-3542326),  [ZBL06621300](https://zbmath.org/?q=an:06621300).</cite>)