$$ C ::= \{ f(n) : \forall n \in N \}$$
is a shorthand for
$$C ::= \{ m \in M : \exists n\in N \text{ such that } m = f(n) \}$$
where $M$ is the codomain of $f$. Here your predicate $\phi(m)$ is $\exists n\in N $ such that $ m = f(n)$.
This abuse of notation is universal: people do often write things like $\{ f(n) : \forall n \in N \}$ without defining what it means, and depend on the reader to understand what is meant. But it's also no trouble to say that if $f$ is a function with domain $A$ and codomain $B$, and $A'\subset A$, then $\{ f(a) : a\in A' \}$ is defined to mean the same as $\{ b\in B : \exists a\in A' \text{ such that } b=f(a) \}$.
Similarly, one often defines a set of ordered pairs using a notation like
$$D ::= \{ \langle p, q\rangle : p\in P, q\in Q, \phi(p,q)\}$$
when what is really meant is
$$D ::= \{z \in P\times Q : \exists p\in P \text{ and }\exists q\in Q\text{ such that } z = \langle p, q\rangle \text{ and }\phi(p,q)\} $$
and it rarely or never seems to cause confusion.
In my experience this is rarely spelled out, and you are very observant to notice it. The first place I saw it discussed explicitly was in the appendix to John L. Kelley's Topology, which first defines the composition of two relations $r$ and $s$:
57 DEFINITION $r\circ s = \{u : \text {for some $x$, some $y$, and some $z$, $u=\langle x,z\rangle,\langle x,y\rangle\in s \text{ and } \langle y,z\rangle \in r$}\}$
and then explains the shorthand:
To avoid excessive notation we agree that $\{\langle x, y\rangle : \cdots\}$ is to be identical with $\{u: \text{ for some $x$, for some $z$, $u=\langle x,z\rangle$, and }\cdots \}$. Thus $r\circ s = \{\langle x,z\rangle : \text{for some $y$, $\langle x,y\rangle\in s$ and $ \langle y,z\rangle \in r$ } \}$.
(1955 edition, page 260.)
Alternatively, consider the following:
For a function $f:Z_+\rightarrow Z_+ $ such that $\exists N \in Z_+ : \forall n > N, f(n) = f(N) $
Let $M = min\{ N : \forall n > N, f(n) = f(N) \}$
and $S(f) = \sum_{i=1}^{M} f(i) + M$
Then for any value $k$, there are only finitely many such functions with $S(f) = k$. List all $f$ with $S(f)=1$ then all $f$ with $S(f)=2$ etc.
Best Answer
Since your domain seems to be fixed throughout your argument, there is no need to make it visible to your reader. In these cases, it's quite common to write $c_x$ for the constant function $c_x \colon D \to \{x\}$, where $D$ is the fixed domain. Then $$ \mathcal C = \{ c_x \mid x \in X \} $$ is the collection of all constant functions with value in $X$ - for some given $X$. In your particular case, $$ \mathcal C = \{ c_z \mid z \in \mathbb Z\} $$ is the collection of all constant functions with integer values.