Quoting Hungerford's Algebra:
A nonzero element $a$ of a commutative ring $R$ is said to divide an element $b \in R$ (notation: $a \mid b$) if there exists $x \in R$ such that $ax = b$.
Let $X$ be a nonempty subset of a commutative ring $R$. An element $d \in R$ is a greatest common divisor of $X$ provided:
- $d \mid a$ for all $a \in X$;
- $c \mid a$ for all $a \in X$ $\implies$ $c \mid d$.
In particular, the greatest common divisor (from now on I use "gcd") of any nonempty subset of a commutative ring $R$ must be a nonzero element of $R$ (else, writing $d \mid a$ would make no sense). Now what's the gcd of $\{0\}$ in an integral domain? Condition 1 is satisfied by every element $d \in R$. In fact, given $d \in R$, we have that $d0 = 0$ for all $d \in R$, hence $d \mid 0$ by definition. What can we deduce from condition 2?
Since every element $c \in R$ satisfies the condition $\forall a \in X:c\vert a $, condition 2 tells us that $d \in R$ needs to be an element such that $\forall c \in R: c \vert d$. There exists only one such element in $R$.
Also, the first statement after the definition is false. In fact, it is precisely what answers your question.According to the definition of Hungerford, a GCD of a subset is indeed defined to be non-zero since a divisor is defined to be non-zero. That given, it is sensible to simply define the GCD of $\{0\}$ as $0$. Note, however, that the definition varies among authors as is also mentioned here.