A dipole consisting of charges $q$ and $-q$ separated a distance $l$ undergoes oscillatory motion because the torque is always trying to restore equilibrium. The magnitude of the torque is given by
\begin{equation}
\tau = -\left| \mathbf p\times \mathbf E \right| = -qlE \sin \theta
\end{equation}
where $\mathbf p$ is the dipole moment and $\mathbf E$ is the electric field. The second law of Newton for rotation gives
\begin{equation}
\tau = I \ddot \theta,
\end{equation}
where $I = \frac{1}{2}ml^2$ is the moment of inertia of the dipole for charges with equal mass. We obtain
\begin{equation}
\ddot \theta + \frac{2qE}{ml} \sin \theta = 0,
\end{equation}
which is equivalent to the equation of motion of a simple gravity pendulum. However, in general, the solution $\theta(t)$ cannot be written down in terms of elementary functions.
Small-angle approximation
For small angles $\theta \ll 1$, the motion is harmonic:
\begin{equation}
\ddot \theta + \frac{2qE}{ml} \theta = 0 \quad \Rightarrow \quad \theta(t) = \theta_0 \cos \omega t, \quad \omega =\sqrt{\frac{2qE}{ml}},
\end{equation}
with $\theta_0 \ll 1$ the initial deviation and where we assumed that the initial angular velocity is zero for simplicity. In case $\mathbf E = E \hat{\mathbf x}$, the dipole moment becomes
\begin{equation}
\mathbf p(t) = p_0 \left( \cos \theta(t) \, \hat{\mathbf x} + \sin \theta(t) \, \hat{\mathbf y} \right) \simeq p_0 \left( \hat{\mathbf x} + \theta_0 \sin \omega t \, \hat{\mathbf y} \right).
\end{equation}
Friction and radiation loss
So far we have ignored the effect of friction for a dipole placed in some medium or the inevitable energy loss from radiation since the charges are accelerated. However, this is different from standard dipole radiation where the dipole moment has a fixed direction but an oscillating magnitude. The average radiated power of a free rotating dipole with fixed magnitude goes as $\omega^4$. A higher $\omega$ corresponds to faster accelerating charges and more radiation loss.
Conclusion
So in conclusion, small deviations from equilibrium result in harmonic motion of the dipole moment around equilibrium. For larger deviations, the motion is still oscillatory but not harmonic (the frequency will depend on the amplitude). If the initial velocity is large enough, the motion is circular (for motion outside of the separatrix in phase space). In this case, the dipole is rotating instead of oscillating.
However, all motion is damped even in absence of friction due to dipole radiation. For circular motion the power loss goes initially as the fourth power of the frequency.
The torque $ \tau $ on an electric dipole with dipole moment p in a uniform electric field E is given by $$ \tau = p \times E $$ where the "X" refers to the vector cross product.
Ref: Wikipedia article on electric dipole moment.
I will demonstrate that the torque on an ideal (point) dipole on a non-uniform field is given by the same expression.
I use bold to denote vectors.
Let us begin with an electric dipole of finite dimension, calculate the torque and then finally let the charge separation d go to zero with the product of charge q and d being constant.
We take the origin of the coordinate system to be the midpoint of the dipole, equidistant from each charge. The position of the positive charge is denoted by $\mathbf r_+ $ and the associated electric field and force by $\mathbf E_+$ and $ \mathbf F_+$, respectively. The notation for these same quantities for the negative charge are similarly denoted with a - sign replacing the + sign.
The torque about the midpoint of the dipole from the positive charge is given by
$$ \mathbf \tau_+ = \mathbf r_+ \times \mathbf F_+ $$
where
$$ \mathbf F_+ = q\mathbf r_+ \times \mathbf E_+(\mathbf r+) $$
Similarly for the negative charge contribution
$$ \mathbf \tau_- = \mathbf r_- \times \mathbf F_- $$
where
$$ \mathbf F_- = -q\mathbf r_- \times \mathbf E_-(\mathbf r-) $$
Note that
$$ \mathbf r_- = -\mathbf r_+ $$
We can now write the total torque as
$$ \mathbf \tau_{tot} = \mathbf \tau_- + \mathbf \tau_+ =q\mathbf r_+ \times (\mathbf E(\mathbf r_+)+\mathbf E(\mathbf r_-))$$
It is clear that in taking the limit as the charge separation d goes to zero, the sum of electric fields will only contain terms of even order in d.
Noting that $$ \mathbf |r_+| = \frac{d}{2} $$
and defining in the usual way $$ \mathbf p = q\mathbf d = q(\mathbf r_+ - \mathbf r_- ) $$
We can write that $$ \tau_{tot} = \mathbf p \times \mathbf E(0) + \ second \ order \ in \ d $$
As we take the limit in which d goes to zero and the product qd is constant, the second order term vanishes.
Thus, for an ideal (point) dipole in a non-uniform electric field, the torque is given by the same formula as that of a uniform field.
Note that it is not correct to start with the expression for a force on an ideal/point dipole in a non-uniform field and then calculate torque from this force. To derive this expression one ends up first taking the limit of a point dipole (on which there is zero force in a uniform field) and then one finds a torque of zero, which is incorrect. One must start with the case of a finite dipole, calculate torque and only then pass to the limit.
When p and E are parallel and anti-parallel, the torque is zero, so yes zero is possible. But the case in which p and E are anti-parallel is one of an unstable equilibrium, and a small angular perturbation will cause the dipole to experience a torque which attempts to align the dipole with the electric field.
Best Answer
If the angle between the field and the dipole moment vector is maintained at $180^0$,the field cannot align the dipole. The field has to exert a torque on the dipole to rotate it and align it along the field direction. Torque can be exerted only if the field can exert a force. But in the position mentioned the field exerts no force on the dipole. But if the dipole is slightly displaced from its unstable equilibrium position, then the field aligns the dipole along the field direction.