Size of an activation record $= 2 + 2 = 4$ bytes.
So, no. of possible activation records which can be live at a time $= 64/4 = 16$.
So, we can have $16$ function calls live at a time (recursion depth $= 16$), meaning the maximum value for $n$ without stack overflow is $16$ (calls from $1-16$). For $n=17$, stack will overflow.
This is different from the total no. of recursive calls which will be as follows:
$$\begin{array}{|c|c|} \hline \textbf{n} & \textbf{No. of calls} \\\hline 1 & 1 \\\hline 2 & 3 \\\hline 3 & 5 \\\hline 4 & 9 \\\hline 5 & 15 \\\hline 6 &25 \\\hline \end{array}$$