Does my AMD-based machine use little endian or big endian?

All x86 and x86-64 machines (which is just an extension to x86) are little-endian.

You can confirm it with something like this:

#include <stdio.h>
int main() {
    int a = 0x12345678;
    unsigned char *c = (unsigned char*)(&a);
    if (*c == 0x78) {
       printf("little-endian\n");
    } else {
       printf("big-endian\n");
    }
    return 0;
}

"Intel-compatible" isn't very precise.

Intel used to make big-endian processors, notably the StrongARM and XScale. These do not use the IA32 ISA, commonly known as x86.

Further back in history, Intel also made the little-endian i860 and i960, which are also not x86-compatible.

Further back in history, the prececessors of the x86 (8080, 8008, etc.) are not x86-compatible either. Being 8-bit processors, endianness doesn't really matter...

Nowadays, Intel still makes the Itanium (IA64), which is bi-endian: normal operation is big-endian, but the processor can also run in little-endian mode. It does happen to be able to run x86 code in little-endian mode, but the native ISA is not IA32.

To my knowledge, all of AMD's processors have been x86-compatible, with some extensions like x86_64, and thus are necessarily little-endian.

Ubuntu is available for x86 (little-endian) and x86_64 (little-endian), with less complete ports for ia64 (big-endian), ARM(el) (little-endian), PA-RISC (big-endian, though the processor supports both), PowerPC (big-endian), and SPARC (big-endian). I don't believe there is an ARM(eb) (big-endian) port.

An easy way to know the endiannes is listed in the article Writing endian-independent code in C

const int i = 1;
#define is_bigendian() ( (*(char*)&i) == 0 )

Assuming you have Python installed, you can run this one-liner, which will print "little" on little-endian machines and "big" on big-endian ones:

python -c "import struct; print 'little' if ord(struct.pack('L', 1)[0]) else 'big'"