Is a matrix with $1$s in the diagonal and off-diagonal entries with absolute value less than $1$ invertible?

well, no. With $n$ by $n,$ and $n \geq 3,$ let all the off-diagonal entries be $$ \frac{-1}{n-1}. $$ So the vector with all entries 1 is an eigenvector with eigenvalue 0.

As Will points out, this isn't true. But you can salvage this idea using the weaker notion of diagonal dominance: if $A$ has 1s along the diagonal and $$\sum_{j\neq i} |a_{ij}| < 1,$$ then $A$ is invertible (the Levy–Desplanques theorem). Notice that for 2x2 matrices, this condition is just that the off-diagonal entries have magnitude less than 1.

The 2x2 case works because it is diagonally dominant. So to explain what happened here, the off-diagonal entries EACH having magnitude less than one isn't enough. They need to be small enough. So if the sum of the absolute values of the off-diagonal elements in a row is strictly less than one and the diagonal-entry has magnitude one, then the matrix will be invertible. Gershgorin proved it so.