@DaveBenson, has already given a beautiful answer to this question. I just wanted to point out that a number of the things he says (although not his full computation of the dimension of the Ext-algebra in the comments) can be deduced from the general theory of semigroup algebras.

A finite semigroup $S$ is nilpotent if it has an absorbing element $0$ such that $S^k=0$ for some $k\geq 1$. The nonidentity elements of your nilCoxeter monoid form a nilpotent semigroup and the smallest $k$, as pointed out by @DaveBenson, is the length of the longest word plus 1. If $M=S\cup \{1\}$, then the contracted monoid algebra $K_0M$ is the $K$-algebra with basis $M\setminus \{0\}$ and product defined by extending the product of $M$, but where the zeroes of $M$ and $K$ are identified.

In this set up, there is always one simple module $L$, the one-dimensional representation sending $1$ to $1$ and $S$ to $0$. The radical of $K_0M$ is the $K$-span of $S$ and therefore the Loewy length is the smallest power $k$ with $S^k=0$. The quiver of $K_0M$ has a single vertex and the number of loops is the number of elements of $S\setminus S^2$, which is also the dimension of $\mathrm{Ext}^1(L,L)$. For the nilxCoxeter monoid, this is the set of $x_s$ with $s$ a Coxeter generator, and so has size the rank of $S$ as pointed out by @DaveBenson. All this is a elementary but is a special case of results than can be found in my book on the representation theory of finite monoids (for instance the quiver computation follows because this is a special case of a $\mathcal J$-trivial monoid and the quiver is computed for such in Chapter 17, but the nilpotent semigroup case is much easier). Note that quiver presentation for $K_0M$ is obtained by taking a semigroup presentation of $S$ with respect to the minimal generating set $S\setminus S^2$. For the nilCoxeter monoid this is well known to be obtained from the Coxeter presentation of $W$ by replacing each relation $s^2=1$ by $x_s^2=0$. So for $S_3$, this is generators $a,b$ and relations $a^2=b^2=aba-bab=0$.

The fact that the algebras is Frobenius can be seen also from the general theory of such semigroups. There is a general criterion due to Wenger that can be found in Section 5 of my paper Factoring the Dedekind-Frobenius determinant of a semigroup. It is easy to see that if $S^k$ is the smallest power which is $0$, then each element of $S^{k-1}$ is in the socle. So to have a chance to be Frobenius (as the algebra has a single one-dimensional simple), one needs $S^{k-1}$ to be a singleton $\{z_0\}$. In your case of the nilCoxeter monoid, $k-1$ is the length of the longest element $w_0$ and $S^{k-1}=\{x_{w_0}\}$. But this is not enough to be Frobenius. Wenger says take the $M\setminus \{0\}\times M\setminus \{0\}$ matrix $A$ where $A_{s,t}$ is $1$ if $st=z_0$ and $0$ else, and the algebra $K_0M$ is Frobenius with socle $Kz_0$ iff the determinant of $A$ is nonzero. This matrix gives the isomorphism of the left regular representation with the dual of the right regular representation (if there is such an isomorphism).

In your case, this matrix is $W\times W$ and the entry $A_{x_u,x_v}$ is $1$ is $uv=w_0$ and $0$ else. This is a permutation matrix since for each $u$ there is a unique $v$ with $uv=w_0$ and $\ell(u)+\ell(v)=\ell(w_0)$ by Coxeter group theory. Therefore, the algebra is Frobenius with socle $Kw_0$.