Consider the data of $f(x)$ given in the table.
\begin{array}{|c|c|c|c|}
\hline i & 0 & 1 & 2 \\
\hline x_{i} & 1 & 2 & 3 \\
\hline f\left(x_{i}\right) & 0 & 0.3010 & 0.4771 \\
\hline
\end{array}
The value of $f(1.5)$ estimated using second-order Newton's interpolation formula is $\_\_\_\_\_\_$. (rounded off to $2$ decimal places).