rol ax, 1 works on bits in that register, not bytes in an array. What you're looking for is like C++ std::rotate(first, new_first, last) to rotate array elements.
You can swap two adjacent bytes in memory with rol word [mem], 8, or even do a dword rotate by 8, 16, or 24 bits to re-arrange bytes in a 4-byte chunk, but that's all you can do with one rotate.
So for example you could do this:
;; starting with prime = .byte 2,3,5,7,11
ror dword ptr [prime], 8 ; 3,5,7,2, 11 ; x86 is little-endian
ror word ptr [prime+4], 8 ; 3,5,7, 11,2
That's GAS .intel_syntax; I don't recognize the syntax you're using. It looks like a hybrid of AT&T and Intel with round parens for addressing modes, but it does seem to be destination on the left.
That's fairly inefficient compared to just loading all 5 bytes and storing them back where you want them. Specifically, you'll get a store-forwarding stall when the word load part of the 2nd ror reads 1 byte that was just stored by the dword store part of the first ror. (A memory-destination instruction decodes to load + ALU + store operations. The first store will still be in the store buffer when the 2nd rotate's load executes. Assuming a modern x86 CPU executing this code natively; but since you're in DOSBox it's actually getting emulated not executed).
The more straightforward way to do this would be:
mov eax, [prime+1] ; dword load: 3,5,7,11
mov dl, [prime] ; byte load: 2
mov [prime], eax
mov [prime+4], cl ; 3,5,7,11,2
This is easy to extend for larger array sizes. If needed you can write a loop to copy elements toward the front while you have the original first element saved somewhere else (in a register).
