Why is the dimension of the vector space which contains only the zero vector $0$?

That means the basis is an empty list. I can see how the empty list spans the space (Basically, linear combinations of nothing are equal to $0$) but I don't see what it means for the empty list to be linearly independent. This is the definition of linear independence I am familiar with:

If $a_1v_1+...+a_mv_m = 0$ then $a_1 = ... = a_m = 0$

That definition does not seem to be applicable for an empty list of vectors. I assume you solve that problem with a better definition of linear independence.

Bonus: Is there a name for the vector space that contains only the zero vector?

Your definition omitted some crucial fine print. A set $B$ in a vector space is linear independent if: "For every subset $\{v_{1}, \dots, v_{m}\} \subset B$, if $\sum_{i} a_{i} v_{i} = 0$ for some scalars $a_{i}$, then $a_{i} = 0$ for all $i$." If $B = \emptyset$, this condition is vacuous.
A vector space with one element is sometimes called a trivial vector space, but often one simply denotes it by "$0$" without giving it a name.