We have a symmetric bilinear form on $\mathbb R^n$ defined by
$$e_1\cdot e_1=e_2\cdot e_2=e_3\cdot e_3=\cdots=0$$
$$e_1\cdot e_2=e_1\cdot e_3=e_2\cdot e_3=\cdots=1.$$
The signature is the numbers of vectors in an orthogonal basis squaring to $+1,-1,0$.
For $n=1$, this is just $e_1\cdot e_1=0$, so the signature is $(0,0,1)$.
For $n=2$, we can find an orthogonal basis $(e_1\pm e_2)/\sqrt2$ which shows that the signature is $(1,1,0)$.
For $n=3$, my answer here shows that the signature is $(1,2,0)$.
Would the general case for $n>1$ have $(1,n-1,0)$ ?