The answer is $f(n)=1+(n-1) \cdot \left\lceil \frac{n-1}{2}\right\rceil$. Indeed, note that one comon difference must have absolute value at least $\left\lceil \frac{n-1}{2}\right\rceil$ and hence two numbers in that row must differ by at least $(n-1) \cdot \left\lceil \frac{n-1}{2}\right\rceil$, hence the maximal number must be at least $1+(n-1) \cdot \left\lceil \frac{n-1}{2}\right\rceil$. On the other hand, it is easy to construct examples: For odd $n$, take $a_{ij}=ij$ for $-\frac{n-1}{2} \le i,j \le \frac{n-1}{2}$ so that all entries are between $-\frac{(n-1)^2}{4}$ and $\frac{(n-1)^2}{4}$ and the AP condition is satisfied. Now just add $1+\frac{(n-1)^2}{4}$ to all numbers to make them positive. This does not change anything about the common differences, of course. For even $n$, take the same construction $a_{ij}=ij$ for $-\frac{n-2}{2} \le i,j \le \frac{n}{2}$ so that all entries are between $-\frac{n(n-2)}{4}$ and $\frac{n^2}{4}$ and the AP condition is satisfied. And now add $1+\frac{n(n-2)}{4}$ to make them all positive and at most $1+\frac{n^2}{4}+\frac{n(n-2)}{4}=1+\frac{n(n-1)}{2}$ as desired.