Is CoStandard Deviation a thing? So there's Standard Deviation, Variance, and Covariance, but is there a co standard deviation? 
If not why not? Is there a fundamental mathematical reason or is it just convention?
If so why is it not used more, or at least really hard to find using Google searches?
I don't mean this to be a flippant question, I'm trying to really question statistics rather than just memorize a bunch of formulas.
 A: The question seems back-to-front. In mathematics we don't invent names for quantities "just because we can", but because the named quantity is useful for something.
The OP's question doesn't give and reasons why he/she thinks there is a useful quantity that might be named "coStandard Deviation" and the answers are guessing at things that might be useful.
To generalize the concept to multi-variable linear regression with $n$ variables, the "covariance" becomes an $n \times n$ symmetric matrix. You can certainly make a sensible definition of the "square root of a symmetric matrix" so long as it is positive definite or semi-definite, but it's hard to think of a use for it in this context - and it isn't the same as taking the square root of each term of the matrix separately! 
Of course the square root of a diagonal matrix (e.g. the variance matrix) is just the square root of the individual terms, so the concept of "standard deviation" does generalize in an obvious and useful way - but "coStandard Deviation" doesn't, IMO. And in general, the "square root of a matrix" isn't even uniquely defined, so which particular square root do you want to choose as the coStandard Deviation?
A: Covariance can be both positive and negative.
So the square root of the covariance could be real or imaginary. 
You can compare a real number with an imaginary number for size. The units for "standard co-deviation" would be inconvenient. There's no benefit in taking the square root. 
A: One useful property of the standard deviation is that it has the same units as the mean, so the magnitudes of $\sigma_X$ and $\bar X$ are directly comparable.  I've never seen anyone compute the co-standard deviation (by which I assume you mean the square root of the covariance); if the units of $X$ and $Y$ are denoted as $[X]$ and $[Y]$, then the units of the covariance are $[X][Y]$ and the units of the co-standard deviation would be $\sqrt{[X][Y]}$, which isn't particularly useful (unless $X$ and $Y$ have the same units). On the other hand, the correlation $\sigma_{XY}/(\sigma_X \sigma_Y)$ is unitless, and is a very common scale for reporting associations.
The variance (in contrast to the standard deviation) is useful because it generally has nicer mathematical properties; in particular
$$
\sigma^2_{X+Y} = \sigma^2_X + \sigma^2_Y + 2 \sigma_{XY},
$$
which simplifies nicely when $X$ and $Y$ are independent (hence $\sigma_{XY}=0$).
While you're thinking about ways of scaling variances you could also consider the coefficient of variation $\sigma_X/\bar X$ (which is unitless), or the variance-to-mean ratio $\sigma^2_X/\bar X$ (which has weird units but is meaningful in the context of a count distribution such as the Poisson, which is also unitless).
