At the request of ShreevatsaR commenting above, I have copied my answer from MO below.
It follows from the Knuth-Netto formula that the asymptotics of $I_n(k)$, for $k$ fixed, is $n^k/k!$ to first order.
I claim the asymptotic behaviour of $I_n^{\sigma(y)=x}(k)$ is as $n^{k-|x-y|}/(k-|x-y|)!$.
For convenience I am going to assume that $x\geq y$. The other case can be treated the same way.
First, we construct a set of permutations of the desired order inside $I_n^{\sigma(y)=x}(k)$. Consider permutations such that
- $\sigma(j)=j$ for $1\leq j\lt y$,
- $\sigma(y)=x$,
- $\sigma(j)=j-1$ for $y+1\leq j \leq x$.
I.e., I have fixed the behaviour of the permutation on $1,\dots,x$, and this has introduced $x-y$ inversions. Now determine the rest of $\sigma$ by choosing an arbitrary permutation on $x+1,\dots,n$ with $k-(x-y)$ inversions. This gives us enough permutations in $I_n^{\sigma(y)=x}(k)$.
Now we show that there are no more than this many (to first order).
It's helpful to think of permutations of $n$ as planar matchings between a top row of $n$ dots and a bottom row of $n$ dots. Then inversions are just points where two arcs cross. (We assume the diagram is drawn in a sensible way so three arcs don't cross at a point, and arcs which don't need to cross, don't cross.)
When looking at permutations which satisfy that $\sigma(y)=x$, we have fixed one arc. If we erase that arc, the resulting diagram has $n-1$ dots left on top and bottom, so it can be viewed as a permutation in $S_{n-1}$.
Suppose we take a permutation in $I_n^{\sigma(y)=x}(k)$ and remove the arc from $y$ to $x$. This arc must be crossed by at least $x-y$ arcs (because it has that many more starting points than ending points to its right). Therefore, erasing this arc removes at least $x-y$ inversions. Thus, we get an injective map
$$I_n^{\sigma(y)=x}(k) \rightarrow I_{n-1}(k-(x-y)) \cup I_{n-1}(k-(x-y)-1) \dots .$$
By the asymptotics already mentioned, the second and later terms on the RHS are lower order, and the first term has the desired asymptotics. This shows that the claimed formula is an upper bound to first order as well.
Consider the following method of building a permutation. For each number in $\{1,2,\dots,n\}$ in increasing order, insert it anywhere in the line of the previously made numbers. The $i^{th}$ number can be inserted in $i$ places, and the number of inversions that insertion will add is anywhere between $0$ and $i-1$. For example, there is only one place to put $1$, effectively creating the list. There are $2$ places to put $2$, either before or after the $1$.
How many ways are there to end up with two inversions? There are two ways this can happen; either there were two steps which each created one inversion (and every other step made zero), or there was only one step which created two inversions. In the former case, there are $\binom{n-1}2$ ways to choose the two steps which created the inversions (as the first step, where $1$ is inserted into an empty list, cannot create any inversions), and there is are $n-2$ ways to choose a step where two inversions are created (as this cannot happen on the first or second steps). Therefore, the total number of ways is
$$
\binom{n-1}2+n-2 = \frac{(n+1)(n-2)}2
$$
Best Answer
Pick $\sigma_1$. That determines that there will be $\sigma_1-1\in\{0,\dotsc,n-1\}$ inversions involving $\sigma_1$, no matter what other values you will pick. The options you have for inversions involving only the remaining values don't depend on the value of $\sigma_1$ you picked; you can imagine the remaining values "compressed" into the range $\{2,\dotsc,n\}$ by moving the values below $\sigma_1$ up by one to fill the gap; that doesn't change the ordering of the values. So the options remaining are the ones for a permutation of $n-1$ objects. Thus the formula follows by induction, since the options $0,\dotsc,n-1$ for $\sigma_1-1$ add a factor $1+x+\dotso+x^{n-1}$ to the generating function for $S_{n-1}$.