You can figure out a way to use vector notation,
x = 1:100;
y = sqrt(x(1:end-3).^2+x(4:end).^2);
I ran this vs the for loop version you have above, and this performed twice as fast when x = 1:1000000 and three times faster when x = 1:100000000. Although it didn't take long for either, I'd be more concerned for memory than speed - but that's only regarding this problem.
Hope this helps!