Prove that the number of answers for $|a_1|+|a_2|+...+|a_k| \le n$ is equal to the number of answers for $|a_1|+|a_2|+...+|a_n| \le k$.

We count the number of solutions to $\sum_{i=1}^n |a_i|\leq k$.

There will be $j\leq r:={\rm min}\{n,k\}$ of the $a_i$ that are $\ne0$, call them $a_{i_1}$, $\ldots$, $a_{i_j}$. The $i_l$ $(1\leq l\leq j)$ can be chosen in ${n\choose j}$ ways. For the chosen $i_l$ put $|a_{i_l}|=x_l+1$ with $x_l\geq 0$. Then we have to count the number of solutions to $x_1+\ldots+x_j+c=k-j$. This number is ${k\choose j}$. It has to be multiplied by $2^j$ in order to account for the choice of the signs of the corresponding $a_{i_l}$. The total number of solutions therefore comes to $$\sum_{j=0}^r{n\choose j}{k\choose j}2^j\ ,$$ which is symmetric in $n$ and $k$.

Let $f(n,k)$ be the number of solutions to $|a_1|+\dots+|a_n|\le k$. Note that $$ f(n,k) = f(n-1,k)+2f(n-1,k-1)+2f(n-1,k-2)+\dots+2f(n-1,0),\tag{*} $$ by conditioning on the value of $a_n$. Rearranging, and using $(*)$ applied to $f(n,k-1)$, we get $$ \begin{align} f(n,k) &= f(n-1,k)+f(n-1,k-1)+[f(n-1,k-1)+2f(n-1,k-2)+\dots+2f(n-1,0)]\\ &= f(n-1,k)+f(n-1,k-1)+f(n,k-1) \end{align} $$ The above shows that $f(n,k)$ obeys a recurrence which is symmetric in $n,k$. Since you can easily verify the symmetric base cases $f(n,0)=f(0,k)=1$, the result $f(n,k)=f(k,n)$ follows by induction.