It is truly a very nice question, one of those questions with an answer one feels must be right, but it is not so clear at first how to prove it.

Nevertheless, aiming at partial progress, I claim that there can be no $\vec\forall\vec\exists$ definition of continuity that works in all real-closed fields, meaning a definition with quantifier complexity $\forall x_0\forall x_1\cdots\forall x_n\exists y_0\exists y_1\cdots \exists y_k\varphi(\vec x,\vec y)$. 

The basic reason that there can be no $\vec\forall\vec\exists$ characterization of continuity is that continuity is not preserved to limits of chains of models. To see this, let us start with the real field $\langle\newcommand\R{\mathbb{R}}\R,+,\cdot,-,0,1,<\rangle$, and build an elementary tower of hyperreal models over it, each with more infinitesimals with respect to the previous. 
 $$\R\prec\R^*_1\prec \R^*_2\prec\cdots$$
Each of these is a real-closed field with the same theory as the real field, and also for the union of the elementary chain $\R^*_\omega$. 

Now, begin with the constant zero function $f_0(x)=0$ in the bottom field (the reals). In each hyperreal field $\R^*_n$, let $f_n$ extend the previous function, still mostly zero, except that we add a new continuous narrow spiking bump from $0$ up to $1$ and back down to $0$ in the new infinitesimal region of $\R_n$ with respect to the previous model. 

Thus, each $f_n$ adds one more flashing bump up to $1$ and back down in the new infinitesimal region of $\R_n$, and $f_n$ has $n$ such bumps. All the functions have $f_n(0)=0$. 

When we expand the language to include these functions, we get a chain of models. 
 $$\langle \R,+,\cdot,-,0,1,<,f_0\rangle\subseteq 
\langle \R^*_1,+,\cdot,-,0,1,<,f_1\rangle\subseteq\cdots$$
Let $\R^*_\omega$ be the union of the fields, with the limit function $f$. 

Notice that although each $f_n$ was continuous in the $n$th model, nevertheless the limit model does not think the limit function $f$ is continuous, since it has $f(0)=0$ but there are bumps up to $1$ arbitrarily close to $0$. The limit model function is discontinuous.

In short, you can always add one more big bump near zero while staying continuous, but the limit model will not think the limit $f$ is continuous, since it has those jumps up to 1 arbitrarily close to 0.

It follows that the property of continuity is not preserved by unions of chains, and so it cannot be characterized by a $\vec\forall\vec\exists$ property, since such kind of properties always are preserved to limits of chains.

**Remark on the underlying theory.** The argument shows that there is no $\vec\forall\vec\exists$ definition that works in all real-closed fields. Emil mentions in the comments, however, that we should be using a stronger theory, namely, the theory $T$ that is true in all structures of the form $\langle\R,+,\cdot,-,0,1,<,f\rangle$ for any choice of function $f:\R\to\R$. That is, what we naturally want is a characterization that works in all these structures. The argument I have given does not quite show that there is no $\vec\forall\vec\exists$ definition of continuity in these structures, since perhaps such a definition works in all these models, but not in all real-closed fields. Since my functions $f_n$ can be taken as definable, each of the models in the tower I build can be taken to satisfy the common theory, and although the limit model is a real-closed field, there seems little reason to suppose it satisfies the common theory. (Indeed, one can arrange that it does not, by coding some forbidden information into the limit function, a little at a time.)