I do not know the answer to your first question. As for the next two the answer
is positive; one need only slightly modify standard proofs for the usual
polynomial ring:

If $R$ is a graded ring then we define the twisted polynomial ring $R[x]_t$ as
the graded ring generated by $R$ and $x$ with $\deg x=1$ and the relations
$rx=(-1)^{|r|}xr$ where $r$ is homogeneous of degree $|r|$. Your ring is then
$k[x_1]_t[x_2]_t\ldots$ (I assume you meant to have a field as base ring) and in
general an element of $R[x]_t$ can be uniquely written in the form
$\sum_nr_nx^n$. We can now prove that $R[x]_t$ is (left) Noetherian if $R$
is. Let $I\subseteq R[x]_t$ be an ideal and let $I'\subseteq R$ be the ideal of
top terms of $I$. Picking generators for $I'$ and representing polynomials for
them gives a finite number of element of $I$ such that any element of $I$ can be
reduced to a polynomial of fixed degree.

As for finite global dimension by general results it suffices to show that $k$
has a finite resolution. Again we can use induction and assume that $k$ has a
finite resolution as $R$-module and then it suffices to show that $R$ has a
finite resolution as $R[x]_t$-module. This is done by considering $R[x]_te \to
R[x]_t$ with $e$ mapped to $x$. Carrying the induction through gives an explicit
resolution which is a sign-twisted version of the Koszul resolution: In
homological degree $k$ it has an $R[x]_t$-basis $e _{i_1}e_{i_2}\cdots e_{i_k}$ where $1\leq
i_1 < i_2 < \cdots < i_k\leq n$ with $d(e_{i_1}e_{i_2}\cdots e_{i_k})=\sum_r
x_{i_r}e_{i_1}\cdots \widehat{e_{i_r}}\cdots e_{i_k}$ (Look Ma no signs!).