How to calculate $\frac{\partial \mathrm{logdet}({X}_{i\times i})}{\partial X}$? I know that $\frac{\partial \mathrm{logdet}(X)}{\partial X}=(X^{-1})^T$.
Lets denote ${X}_{i\times i}$ which is $i\times i$ matrix that is formed from $X\in\mathbb{R}^{n\times n}$ by removing all columns $i+1,i+2,\ldots,n$ and rows $i+1,i+2,\ldots,n$.
Ex: if $X=\begin{bmatrix}1 & 2  & 3\\4  & 5 &6 \\7  & 8  &9\end{bmatrix}$, then $X_{1\times 1}=1$, $X_{2\times 2}=\begin{bmatrix}1 & 2 \\4  & 5 \end{bmatrix}$.
How to calculate $\frac{\partial \mathrm{logdet}({X}_{i\times i})}{\partial X}$?
 A: Just note that using the chain rule you get
$
\left(\frac{\partial\log(\det(X_{i\times i}))}{\partial X}\right)_{lg}=
\left(\frac{\partial\log(\det(X_{i\times i}))}{\partial X_{i\times i}}\frac{\partial X_{i\times i}}{\partial X}\right)_{lg}$
It remains to compute $\frac{\partial X_{i\times i}}{\partial X}$ and this is easily done considering that
$\left(\frac{\partial X_{i\times i}}{\partial X}\right)_{lj}=\frac{\partial X_{i\times i}}{\partial X_{lj}}=\begin{cases}1\ \ \text{if}\ l,j\le i\\ 0\end{cases}=\begin{pmatrix} &I_{i}&0_{n-i}\\
&0_{n-i}&0_{i}\end{pmatrix}$
Using the result you state at the start of your question we get:
$\frac{\partial\log(\det(X_{i\times I}))}{\partial X}=\begin{pmatrix} &(X_{i\times i}^{-1})^T&0_{n-i}\\
&0_{n-i}&0_{i}\end{pmatrix}$
This is true in general: whenever you consider a function restricted to $X_{i\times i}$, the derivative is going to be: $\frac{\partial f(X_{i\times i})}{\partial X}=\begin{pmatrix} &f'(X_{i\times i})&0_{n-i}\\
&0_{n-i}&0_{i}\end{pmatrix}$
A: Consider a matrix analog of the first cartesian basis vector.
$$\eqalign{
&E &= \pmatrix{I\\{\tt\large 0}} \in {\mathbb R}^{n\times i}\cr
}$$ where $I\in{\mathbb R}^{i\times i}$ is the identity matrix and 
${\tt\large 0}\in{\mathbb R}^{(n-i)\times i}$ is the zero matrix.
This $E$ matrix can be used to extract the leading $(i\times i)$ block-diagonal of the $X$ matrix, i.e.
$$X_{i\times i} = E^TXE \;\doteq\; Y $$
The goal is to calculate the gradient of the function $\,\phi=\log\det(Y)$. 
First calculate the differential, then perform a change of variables from $Y\to X$.
$$\eqalign{
\phi &= \log\det(Y) \\
d\phi
 &= (Y^T)^{-1}:dY \\
 &= Y^{-T}:E^TdX\,E \\
 &= E\,Y^{-T}E^T:dX \\
\frac{\partial \phi}{\partial X} 
 &= E\,Y^{-T}E^T \\
 &= \pmatrix{Y^{-T}&0\\0&0} \\
}$$
In some of the steps above, a colon is used to denote the matrix inner product, i.e.
$$A:B = {\rm Tr}(A^TB)$$
