Why is the cosmological constant a scalar? 
*

*Maybe my understanding is just off, but the cosmological constant is just a scalar, right?

*What are it's units?

*Why a scalar? - was a tensor 'cosmological constant' ever considered or is it just not conformable to the math or physics of the field equations?
 A: Yes it is, its units in the SI system are $m^{-2}$ and you can guess it from a dimensional analysis of the Einstein field equation. From this equation you can also guess that imposing a scalar value is the simplest approach to the problem, there are probably theories and models less simpler than this, but I believe that the scalar value is still used because it was originally proposed by Einstein to impose the existence of a static universe.
The Einstein field equation is:
$R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu} - \Lambda g_{\mu \nu}$
A: As mentioned in another answer, it was first added by Einstein to make the universe static. As to why it is a scalar, I paraphrase from Straumann (2013).

It is natural to postulate that the ten potentials $g_{\mu\nu}$ satisfy equations of the form
  $$\mathcal{D}_{\mu\nu}[g]=T_{\mu\nu}$$
  where $\mathcal{D}$ is a tensor field constructed pointwise from $g$ and its first and second partial derivatives. Furthermore, this tensor field should satisfy
  $$\nabla^\mu\mathcal{D}_{\mu\nu}=0$$
  in order to guarantee $\nabla^\mu T_{\mu\nu}=0$. 
  [Some remarks follow.]
Theorem (Lovelock) A tensor $\mathcal{D}_{\mu\nu}$ with the required properties is in four dimensions 
  $$\mathcal{D}_{\mu\nu}=aG_{\mu\nu}+bg_{\mu\nu}$$
  where $a,b\in \mathbb{R}$.

In four dimensions, the field equations are quite unique. $a=(8\pi G)^{-1}$ is determined by taking the Newtonian limit. $b=\Lambda/8\pi G$ is chosen such that the field equations have the simple form you are used to. In four dimensions, it is not possible to have anything else other than a cosmological constant scalar. 
A more physical interpretation is this: a constant tensor in spacetime would violate cosmological isotropy. It would pick out a preferred direction in a given coordinate system. 
I should note that there exists a generalization of the Lovelock theorem to higher dimensions, but I am unaware of the proof. In any case, the associated length scales of something like Kaluza-Klein extra dimensions must be very small such that we needn't worry about more than 4 dimensions in standard GR. (See any text on string theory for extra-dimensional stuff.)
