I have little intuition about the projective hierarchy, and this is presumably far too simple-minded to come close to a sharp estimate, but for what it's worth: the following easy argument shows that the set $S$ of surjective continuous functions is at worst a $\boldsymbol{\Pi}_{2}^1$-set. I have no idea whether $S$ is Borel or not.
The point is that we can describe $S$ by quantifying twice over a Polish space: for all $y \in Y$ there exists $x \in X$ such that $f(x) = y$ or, equivalently, $(f,x,y) \in F$, where $F$ is the closed set $ F = \{(f,x,y) \in C(X,Y) \times X \times Y\,:\,f(x) = y\} \subset C(X,Y) \times X \times Y. $ The set $F$ is closed because the map $\operatorname{ev}\colon C(X,Y)\times X \to Y$ given by $\mathrm{ev}(f,x) = f(x)$ is continuous because $X$ is locally compact, and $F$ is the pre-image of the diagonal of $Y \times Y$ under $\mathrm{ev} \times \mathrm{id}_Y \colon C(X,Y) \times X \times Y \to Y \times Y$.
Projecting $F$ down to $C(X,Y) \times Y$ (existential quantification over $X$) gives us the $\boldsymbol{\Sigma}_{1}^1$-set (aka analytic set) $ A = \{(f,y) \in C(X,Y) \times Y\,:\,(\exists x \in X)\;f(x) = y\} \in \boldsymbol{\Sigma}_{1}^1. $ Now $\boldsymbol{\Sigma}_{1}^1 \subset \boldsymbol{\Pi}_{2}^1$ and the point class $\boldsymbol{\Pi}_{2}^1$ is stable under universal quantification over a Polish space (here $Y$) — see Kechris, Classical Descriptive Set Theory, Theorem 37.1 on page 314 but this is precisely the set we are interested in: $ \begin{align*} S &= \{f \in C(X,Y)\,:\,(\forall y \in Y)\;(f,y) \in A\} \\ & = \{f \in C(X,Y)\,:\,(\forall y \in Y)\,(\exists x \in X)\;(f,x,y) \in F\} \\ \end{align*} $ so we've shown that $S \in \boldsymbol{\Pi}_{2}^1$.