Does bit-shift depend on endianness?

Endianness is the way values are stored in memory. When loaded into the processor, regardless of endianness, the bit shift instruction is operating on the value in the processor's register. Therefore, loading from memory to processor is the equivalent of converting to big endian, the shifting operation comes next and then the new value is stored back in memory, which is where the little endian byte order comes into effect again.

Update, thanks to @jww: On PowerPC the vector shifts and rotates are endian sensitive. You can have a value in a vector register and a shift will produce different results on little-endian and big-endian.

No, bitshift, like any other part of C, is defined in terms of values, not representations. Left-shift by 1 is mutliplication by 2, right-shift is division. (As always when using bitwise operations, beware of signedness. Everything is most well-defined for unsigned integral types.)

Whichever shift instruction shifts out the higher-order bits first is considered the left shift. Whichever shift instruction shifts out the lower-order bits first is considered the right shift. In that sense, the behavior of >> and << for unsigned numbers will not depend on endianness.