Can there be a magic square with equal diagonal sums different from equal row and column sums?
I assume you are looking for one with not just 'all different numbers', for that is trivial: just take any $n \times n$ magic square and add the same amount (larger than $n^2$) to the cells on the diagonals. For example, you can go from:
\begin{array}{|c|c|c|c|} \hline 16&3&2&13\\ \hline 5&10&11&8\\ \hline 9&6&7&12\\ \hline 4&15&14&1\\ \hline \end{array}
to:
\begin{array}{|c|c|c|c|} \hline 116&3&2&113\\ \hline 5&110&111&8\\ \hline 9&106&107&12\\ \hline 104&15&14&101\\ \hline \end{array}
... and that's just too easy! So, I assume you mean that you have to use all numbers $1$ through $n^2$.
Well, after trying a bunch of things I am fairly convinced that you cannot have a $4 \times 4$ square with numbers $1$ through $16$ where are rows and columns sum p to the same amount (this is actually called a 'semi-magic square') but where the diagonals sum up to the same amount, yet different from the rows and columns. In all $4 \times 4$ semi-magic squares that were not $4 \times 4$ magic squares that I looked at, I found the diagonals still adding up to exactly twice the sum of a row. I don't have a proof though that this is really impossible.
I did, however, find a $6 \times6$ square using numbers $1$ through $36$ with all rows and columns adding up to $111$ but both diagonals adding up to only $97$:
\begin{array}{|c|c|c|c|c|c|} \hline 6&34&1&28&24&18\\ \hline 14&7&36&4&20&30\\ \hline 33&3&31&2&17&25\\ \hline 27&22&23&16&12&11\\ \hline 10&13&5&35&29&19\\ \hline 21&32&15&26&9&8\\ \hline \end{array}
And so, the answer to your question is Yes! (But no, I did not look at the $5 \times 5$ case (although my guess is that you can find an example of what you want for the $5 \times 5$ case and up), and no, I did not create this example in a systematic way: I started with a known semi-magic $6 \times 6$ square and kept swapping rows and columns in a semi-random fashion until I found this one).
Finally, in my research I found that there are many different kinds of magic squares (I had no idea!), such as these 'semi-magic squares' or 'extremely magic squares', or .... but what you are asking about I did not see a name for. Given that they apparently exist you should definitely come up with a name for these!
Aha! As I thought, it also works for $n=5$. Here is one:
\begin{array}{|c|c|c|c|c|} \hline 19&6&15&2&23\\ \hline 9&17&10&13&16\\ \hline 21&24&14&5&1\\ \hline 4&11&8&20&22\\ \hline 12&7&18&25&3\\ \hline \end{array}
Rows and columns sum to $65$, but columns sum to $73$
EDIT 2: Aha! I was wrong about the $4 \times 4$: it is possible to have one!! See Taneli Huuskonen's answer.
I wrote a program to search for $4\times 4$ examples by brute force. Here is one. $$\begin{matrix} 1&11&10&12\cr 3&15&7&9\cr 14&2&13&5\cr 16&6&4&8\cr \end{matrix}$$