Either kind of flow can exhibit steady flow or unsteady flow. Unsteady flow is where, over a large time scale, things are changing at each spatial location with time. When they say that turbulent flow is inherently unsteady, what they mean is that, over a small time scale (at each spatial location), the velocity components are varying rapidly with time, but, when averaged over relatively small time intervals, the average velocity components vary much more slowly, or not at all. If they do not vary at all, the turbulent flow is considered steady. If the vary slowly (at each spatial location) the turbulent flow is considered unsteady.
Regarding inviscid flow, it is assumed that viscous and turbulent stresses are both considered insignificant. So it is considered neither laminar nor turbulent. It would be what you would obtain if the fluid had a very low viscosity, but the flow could not transition to turbulent flow.
There is a hypothesis left over here: the flow is irrotational, i.e. the so-called vorticity
$$\vec{\omega}=\nabla\times\vec{v}$$
is zero. As a result, the velocity field is a gradient,
$$\vec{v}=\nabla\phi,$$
for some scalar field $\phi$: the field appearing in the Bernoulli equation you quoted. Note that the right-hand side is not necessarily zero but it can be taken as a constant, which does not depend either on position or on time.
This is a simple consequence of (i) Euler equation for an incompressible fluid,
$$\frac{\partial\vec{v}}{\partial t} + (\vec{v}\cdot\nabla) v + \nabla\left(\frac{p}{\rho}+\psi\right)=0,$$
where $\psi=gh$ but it could be the potential for any other conservative force field; and (ii) the identity
$$\vec{v}\times\vec{\omega} = \nabla\left(\frac{v^2}{2}\right) - (\vec{v}\cdot\nabla) v.$$
Then Euler equation reads
$$\nabla\left(\frac{\partial\phi}{\partial t}+\frac{v^2}{2}+\frac{p}{\rho}+\psi\right)=0.$$
Therefore the term between parentheses is function of time only but we can absorb any time dependence into $\phi$, and therefore we end up with
$$\frac{\partial\phi}{\partial t}+\frac{v^2}{2}+\frac{p}{\rho}+\psi=C$$
where $C$ is constant over space and time. This is therefore a stronger result than the traditional Bernoulli theorem where the constancy is only along each streamline, with a priori a different constant for each streamline. Here the constancy is over the entire fluid.
Best Answer
Yes, there can be. But, what we know is that in an incompressible flow (as you have defined it), the density field is always steady -- which is obvious because we are defining the density to not change.
You've only listed the conservation of mass equation, but remember there are 2 more -- conservation of momentum and conservation of energy. Incompressibility does nothing to remove the time derivatives from these equations. It does, however, decouple the energy equation from the rest of the system and I encourage you to work through the entire derivation and implications of the complete system.
As a final note, and this is being a bit pedantic, but incompressibility really means that:
$$\frac{\partial \rho}{\partial p} \approx 0 $$
and not that the density is constant. It is entirely possible (and common) to solve the "incompressible" equations where density varies due to multi-species flows (think helium and air, or salt water and fresh water mixing) or due to thermal gradients (hot air is less dense than cold). What you have defined is constant density flow, which is more restrictive than incompressible. The terms are commonly used interchangeably, but it still bothers me enough that I felt I should point it out! Low speed combustion and many atmospheric flows rely on this distinction.