$\begin{bmatrix} 101 & 2 &3 &4 &5 \\ 1 & 102 &3 &4 &5 \\ 1 & 2 & 103 &4 &5 \\ 1& 2 &3 &104 &5 \\ 1 & 2 &3 &4 & 105 \end{bmatrix}$
Sum of all rows of this matrix is 115 hence $\exists \lambda = 115$
Now is there any property that can be used to find out remaining eigenvalues of this matrix?
remaining 4 eigenvalues are 100.
Let $A$ be your matrix and let$$B=\begin{bmatrix}1&2&3&4&5\\1&2&3&4&5\\1&2&3&4&5\\1&2&3&4&5\\1&2&3&4&5\end{bmatrix};$$then $A=B+100\operatorname{Id}$. Since the dimension of the space spanned by the columns of $B$ is $1$ and since $B$ is a $5\times5$ matrix, the characteristic polynomial of $B$ is of the type $x^4(a-x)=ax^4-x^5$, for some $a\in\mathbb R$. On the other hand, by the same argument that you used, $15$ is an eigenvalue of $B$. Therefore, the eigenvalues of $B$ are $0$ ($4$ times) and $15$. So, the eigenvalues of $A$ are $100$ ($4$ times) and $115$.