Phase response of a FIR Hilbert transformer
The slope represents a delay of $N$ samples. You can center the impulse response to time zero by multiplying the transfer function by $e^{i\omega N}$ (transfer function of a negative delay of $N$ samples) or np.exp(1j*w*N) in Python:

from scipy import signal
import numpy as np
N = 10
b = [2*np.sin(np.pi*n/2)**2/(np.pi*n) for n in range(1, N+1)]
b = np.append(-1*np.flip(b, 0), np.append([0], b))*np.hamming(2*N+1)
w, h = signal.freqz(b)
h = h*np.exp(1j*w*N)
import matplotlib.pyplot as plt
fig = plt.figure()
plt.title('Digital filter frequency response')
ax1 = fig.add_subplot(111)
plt.plot(w, 20 * np.log10(abs(h)), 'b')
plt.ylabel('Amplitude [dB]', color='b')
plt.xlabel('Frequency [rad/sample]')
ax2 = ax1.twinx()
angles = np.angle(h)
plt.plot(w, angles, 'g')
plt.ylabel('Angle (radians)', color='g')
plt.grid()
plt.axis('tight')
plt.show()


Most of the above is from Scipy documentation which you link to. This gets plotted:

Figure 1. Frequency response of the Hilbert transformer centered at time zero.

The phase goes to zero at frequency zero, because the phase frequency response transitions from $-\pi/2$ for positive frequencies to $\pi/2$ for negative frequencies.