As pointed by learner, awgn adds noise to the sequence y. So you are not plotting the autocorrelation of white gaussian noise, but the autocorrelation of white gaussian noise plus a constant.
Use zeros instead of ones ; or analyze the difference $z - y$ ; or since $y$ is a constant signal, remove the mean of $z$.
Once point 1 is solved, and since your sequence is of finite length, you will be plotting the autocorrelation of white noise multiplied by a square window. What you will see is thus the autocorrelation of white noise convolved by the autocorrelation of a square window. The autocorrelation of a square window has a triangular shape.
Another way to look at it: the further you move from 0, the less data is present in your input vector. For example, in the sequence:
$-1, 4, 5, 6, -2$
There are 4 pairs of samples distant by a lag of 1, 3 pairs of samples distant by a lag of 2 (-1 and 5 ; 4 and 6 ; 5 and -2), and only two pairs of samples distant by a lag of 3 (-1 and 6, 4 and -2). Thus, there won't be as much data to estimate correctly the autocorrelation function for the larger values of the lag.
You can artificially compensate for the windowing by providing the 'unbiased' argument to xcorr.
Keep in mind that the theory is about infinite length sequences (or for finite length sequences, about the expected value of the result).