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, I claim that there can be no $\vec\forall\vec\exists$ definition of continuity, meaning a definition with quantifier complexity $\forall x_0\forall x_1\cdots\forall x_n\exists y_0\exists y_1\cdots \exists y_k\varphi(\vec x,\vec y)$. This is a sense in which the $\varepsilon\delta$ definition is optimal.

The basic reason 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. 
 $$\R\prec\R^*_1\prec \R^*_2\prec\cdots$$
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.

(Note that all the models $\R^*_n$ are real-closed fields and elementarily equivalent to $\R$ in the language of ordered fields. The chain is not elementary in the language with $f$, of course.)