One can show that $\chi(n,(0))=\dim_K A_n=\binom{N+n}{N}$ since this is the number of monomials of degree $n$. This is a combinatorial statement which is discussed here.
Now, we can choose a graded, free resolution of $M:=A/\mathfrak{a}$
$\cdots\xrightarrow{\quad\phi_2\quad} F_1 \xrightarrow{\quad\phi_1\quad} F_0 \xrightarrow{\quad\phi_0\quad} M \xrightarrow{\quad\quad} 0$ with $ F_i = \bigoplus_{j=1}^{r_i} A(-a_{ij})$ where $A(-a)$ is the ring $A$, degree-shifted by $a\in\mathbb{Z}$, i.e. $A(-a)_n=A_{n-a}$. It is easy to see that $\dim_K A(-a)_n = \binom{N+n-a}{N}$ and it is then an easy calculation using the exactness to show that $\chi(n,\mathfrak{a})=\dim_K M_n = \sum_{i\ge 0} (-1)^i \cdot \sum_{j=1}^{r_i} \binom{N+n-a_{ij}}{N}.$
For $n > \max_{ij} (a_{ij}-N)$, the binomial coefficient $\binom{N+n-a_{ij}}{N} = \frac{(n+N-a_{ij})(n-1+N-a_{ij})\cdots(1+n-a_{ij})}{N!}$ is a polynomial expression in $n$. Let us denote this Polynomial by $P_M(n)$. We now want to show that this polynomial has degree equal to the dimension $d$ of the variety defined by $\mathfrak{a}$.
To do so, we perform induction on the dimension of $M$. If $\dim(M)=0$, then $M$ is Artinian and a finite-dimensional, graded $K$-vector space. This means $P_M(n)=0$ for large enough $n$ and hence, $P_M=0$. In the induction step, we know that $\mathrm{codim}(\mathfrak{a})=N+1-\dim(M)\le N,$ and thus there exists an $x\in A$ not contained in any prime which is minimal over $\mathfrak{a}$. We set $\mathfrak{b}:=\mathfrak{a}+(x)$ and $N:=M/(x)$, so $\mathrm{codim}(\mathfrak{b})=\mathrm{codim}(\mathfrak{a})+1$. Consequently, $\dim(N)=\dim(M)-1$.
The sequence $0\to M(-1)\xrightarrow{\,\cdot x\,} M\to N\to 0$ is exact (in every degree) and this implies $P_M-P_{M(-1)}=P_N$. In other words, $P_N$ is the first difference of $P_M$ and thus, $\deg(P_N)=\deg(P_M)-1$.
Note: The polynomial $P_M$ is usually called the Hilbert Polynomial of the module $M$, while your notion of $\chi(n,\mathfrak{a})$ is often referred to as $H_M(n)$ and called the Hilbert Function. It is important to note that the Hilbert Function itself is not always a Polynomial, but it does agree with $P_M$ for large values of $n$.