This is analogous to the definition of an empty product in mathematics. For a finite non-empty set $S=\{s_1,\ldots,s_n\}$, the product over $S$ can be defined as
$$\prod_{s\in S}s=s_1\times \cdots\times s_n.$$
For such a product you'd want disjoint unions to map into products: if $R\cap S=\emptyset$, then you want $\prod_{x\in R\cup S}x=\left(\prod_{s\in S}s\right) \times \left(\prod_{r\in R}r\right)$, but for this to make sense you want to be able to handle the empty set, and the only way to make the rules consistent is to set
$$\prod_{s\in\emptyset}s=1.$$
This essentially says: if there's nothing to multiply, the result is one. (Similarly, empty sums are defined to be zero, for the same reason.) In the case in hand, you could simply say if there are no units to multiply, then you get one. As Luboš points out, this is the harmless only consistent choice, as multiplying by one does not change the quantity.
Moreover, this empty-product intuition can be carried out to a full formalization of physical dimensions and units as a vector space. The whole works is in this answer of mine, but the essential idea is that positive physical quantities form a vector space over the rationals, where "addition" is multiplication of two quantities and "scalar multiplication" is raising the quantity to a rational power. This vector-space formalism is precisely the reason why dimensional analysis often boils down to a set of linear equations. Moreover, in this vector space the 'zero' is the physical quantity and unit $1$ - neither vector space makes sense unless $1$ is both a quantity and a unit.
Ultimately, of course, it boils down to convention, so people can just say "I'm going to do this in this other way" and they won't be "wrong" as such. However, in general, the consistent way to assign things is to say that dimensionless quantities have dimension $1$ (modulo whatever square bracket convention you're using) and unit $1$.
To back this up a bit, for those that care about organizational guidance, the BIPM publishes the International Vocabulary of Metrology, which states (§1.8, note 1) that
The term "dimensionless quantity" is commonly used and is kept here for historical reasons. It stems from the fact that all exponents are zero in the symbolic representation of the dimension for such quantities. The term "quantity of dimension one" reflects the convention in which the symbolic representation of the dimension for such quantities is the symbol 1 (see ISO 31-0:1992, 2.2.6).
This is essentially the same in the ISO document, which has been superseded by ISO 80000-3:2009 (paywalled, but free preview available), which has an essentially identical entry in §3.8.
Finally, and as a response to some of the comments by Luboš Motl, this applies to the term "physical dimension" as understood by the majority of physical scientists.
There is also an alternative convention, used in high-energy contexts where you work in natural units with $\hbar=c=1$, in which you're left with a single nontrivial dimension, usually taken to be mass (=energy). In that context, it is usual to say a quantity or operator has "dimension $N$" to mean that it has mass dimension $N$ i.e. it has physical dimension $m^N$, but since there's only ever mass as the base quantity it often gets dropped. However, this is very much a corner case with respect to the rest of physical science, and high-energy theorists are remiss if they forget that their "dimension $N$" only works in natural units, which are useless outside of their small domain.
Best Answer
A unit vector has magnitude $1$ - as in, the dimensionless number $1$. Not $1\ \mathrm{cm}$ or $1\ \mathrm{kg}$ or $1\ \mathrm{N}$ or $1\ \mathrm{J}$. It's also not hard to show that for any vector $\vec A$, the dimensions of $\vec A$ and $\vert \vec A \vert$ are the same.