The empty set is indeed a set (the set of no elements) and it is a subset of every set, including itself. $$\forall A: \emptyset \subseteq A,\;\text{ including if}\;\; A =\emptyset: \;\emptyset \subseteq \emptyset$$
$$\text{BUT:}\quad\emptyset \notin \emptyset \;\text{ (since the empty set, by definition, has no elements!)}$$
That is, being a subset of a set is NOT the same as being an element of a set: $$\quad\subseteq\;\, \neq \;\,\in: \;\; (\emptyset \subseteq \emptyset), \;\;(\emptyset \notin \emptyset).$$
$\emptyset \;\subseteq \;\{1, 2, 3, 4, 5\},\quad$ whereas $\;\;\emptyset \;\notin \;\{1, 2, 3, 4, 5\},\;$.
$\{3\} \subseteq \{1, 2, 3, 4, 5\},\quad$ whereas $\;\;3 \nsubseteq \{1, 2, 3, 4, 5\}, \text{... but}\; 3 \in \{1, 2, 3, 4, 5\}$.
An $n$-any function on a set $X$ is a function $X^n\to X$, where $X$ is the $n$-fold cartesian product of $X$ with itself. Thus, a $0$-ary function is a function $X^0\to X$, not $\emptyset \to X$. So the question is to figure out what $X^0$ is. One way to reason about what it should be is to note that $X^n\times X^m$ is essentially the same as $X^{n+m}$ (as most people will agree is true for all $m,n>0$. To make this true also for $n=0$, we need, e.g., that $X^0\times X^m$ is essentially the same as $X^m$. Which set has $Y$ has the property that $Y\times X^m$ is essentially just $X^m$ (for $X\ne \emptyset$)? Well, a moment's thought should reveal that the answer is that $Y$ can be any singleton set.
So, to preserve some basic realizations about the cartesian product of sets, it makes sense to define $X^0$ (for nonempty $X$) to be a singleton set (whichever one you want).
Another way to argue is categorically. The cartesian product of sets is a special case of the notion of categorical product, and $X^0$ corresponds to an empty product. The universal property for the empty product is just a terminal object in the category. The terminal objects in the category of sets are precisely the singletons.
I just saw your edit: Notice that the conventions agree: $X^0=X^\emptyset =\{\emptyset \to X\}$ is a singleton set.
Best Answer
$f$ is a function iff for all $x$ in the domain there exists exactly one $y$ in the codomain such that $(x,y)\in f,$ and in that case we write $f(x)$ for the uniquely identified $y.$ Note the in that case: the notation $f(x)$ presupposes that $x$ belongs to the domain. For the empty function no $x$ can belong to the domain and therefore the notation $f(x)$ cannot acquire meaning.
The notation $f()$ would suggest a unary relationship; functions are by definition binary relationships.