Here is my attempt :
If order of $A$ is $2$, then $$A= \begin{pmatrix} 2 & 2 \\ 2 & 5 \\ \end{pmatrix} \Rightarrow \det A=6.$$ Also the eigenvalues in this case are $1,6$.
If order $A$ is $3$, then $$A= \begin{pmatrix} 2 & 2 & 3 \\ 2 & 5 & 6 \\ 3 & 6 & 10 \\ \end{pmatrix} \Rightarrow \det A=15$$ Also the eigenvalues in these case are $1,1,15.$
If order of $A$ is $4$, then $$ \begin{pmatrix} 2 & 2 & 3 & 4 \\ 2 & 5 & 6 & 8\\ 3 & 6 & 10 & 12\\ 4 & 8 & 12 & 17\\ \end{pmatrix} \Rightarrow \det A=-21.$$ I found that $1$ is an eigenvalue here by directly putting $\lambda=1$ in the characteristic determinant. But $-21$ is not the eigenvalue.
I couldn't use information about eigenvalues so I looked at the pattern of determinants :
$6,15,-21,...?$ Observed that they are $3\times 2,3 \times 5,-3 \times 7$. From this pattern it might be possible that next term is $-3 \times 11$ since $2,5,7$ are primes and next is $11$. Also the negative sign can be in the pattern $+,+,-,-,+,+,-,-,+,+...$ As it seems to me I will have to calculate order $5$ too to get some more concrete idea with respect to this way. But isn't there some elegant method?
I see that every such order $n$ matrix is "engraved to upper-left inside" order $n+1$ such matrix and every $A$ is symmetric. Thanks.
PS : I wrongly calculated $\det A_{4 \times 4}=-21$. It should be $31$.
