Short answer is: The determinant plays a role because it is tied to the jacobian of the multivariate change of variables and the logarithm is tied to taking the log-likelihood.
Long Answer: Let's start with the univariate standard normal density (parameter free) which is $$\frac{1}{\sqrt{2\pi}} \exp\left(-\frac{1}{2}t^2\right).$$
When we extend (parametrize) it to $x=\sigma t + \mu$, the change of variable requires $dt=\frac{1}{\sigma}dx$ making the general normal density
$$\frac{1}{\sqrt{2\pi}} \frac{1}{\sigma}\exp\left(-\frac{1}{2}(\frac{x-\mu}{\sigma})^2\right).$$
Let us also work out the ML estimation of $\mu$ and $\sigma$ simultaneously (when both unknown). The log-likelihood is
$$\text{A constant} - \frac{n}{2}\log(\sigma^2) -\frac{1}{2}\sum_{i=1}^n\left(\frac{x_i-\mu}{\sigma}\right)^2$$ maximization of which is equivalent to minimizing
$$n \log(\sigma^2) + \sum_{i=1}^n\left(\frac{x_i-\mu}{\sigma}\right)^2$$
and both terms involving $\sigma^2$ need to be accounted for in the minimization (with respect to $\sigma^2$).
Multivariate (say number of dimensions = $d$) analogues behave the similar way. Starting with the generating (standard) density
$$\left(\sqrt{2\pi}\right)^{-d} \exp\left(-\frac{1}{2}\mathbf{z}^t\mathbf{z}\right)$$
and the general MVN density is
$$\left(\sqrt{2\pi}\right)^{-d} \left|\boldsymbol{\Sigma}\right|^{-1/2}\exp\left(-\frac{1}{2}\left(\mathbf{x}-\boldsymbol{\mu}\right)^t\boldsymbol{\Sigma}^{-1}\left(\mathbf{x}-\boldsymbol{\mu}\right)\right).$$
Observe that $\left|\boldsymbol{\Sigma}\right|^{-1/2}$ (which is the reciprocal of the square root of the determinant of the covariance matrix $\boldsymbol{\Sigma}$) in the multivariate case does what $1/\sigma$ does in the univariate case and $\boldsymbol{\Sigma}^{-1}$ does what $1/\sigma^2$ does in the univariate case. In simpler terms, $\left|\boldsymbol{\Sigma}\right|^{-1/2}$ is the change of variable "adjustment".
The maximization of likelihood would lead to minimizing (analogous to the univariate case)
$$n \log\left|\boldsymbol{\Sigma}\right| + \sum_{i=1}^n\left(\mathbf{x}-\boldsymbol{\mu}\right)^t\boldsymbol{\Sigma}^{-1}\left(\mathbf{x}-\boldsymbol{\mu}\right)$$
Again, in simpler terms, $n \log\left|\boldsymbol{\Sigma}\right|$ takes the spot of $n \log(\sigma^2)$ which was there in the univariate case. These terms account for corresponding change of variable adjustments in each scenario.
Above is based on taking $\rho(x)=x$ as in the http://arxiv.org/pdf/1206.1386v2 language. Using $\rho(x)=\frac{d}{2}\log x$ (discussed after I.5 on p.2) changes things accordingly (although, as noted in the paper this $\rho(x)$ does not give a valid density).