69 degrees of arc
Apparent size in the sky
The formula for the visual angle of an object is
$$\alpha = 2\arctan{\frac{2r}{2d}},$$
where $\alpha$ is the visual angle of the object, $r$ is the radius of the object (for a sphere, which is what we will be talking about), and $d$ is the distance from the viewer to the object.
Limit of $d$: how close can a moon be?
The Roche limit for a rigid spherical moon, while taking its synchronous rotation into account is
$$d_{roche} = \left(\frac{9M_M}{4\pi\rho_m}\right)^{1/3},$$
where $d_{roche}$ is the Roche limit, $M_M$ is the mass of the planet (Earth in this case, $5.97\times10^{24}$ kg), and $\rho_m$ is the density of the moon.
Limit of $r$: how big can the moon be?
The limit for the size of the moon is the point where the moon becomes more massive than the planet. Thus, the mass of the moon must be, at the most, equal to the mass of the Earth.
There are many ways we can express this mass, but wish to solve for $r$ in terms of something we are already using as a variable; namely, $\rho_m$. Thus
$$\begin{align}M_M &= M_m \\
&= \frac{4}{3}\pi r^3\rho_m \\
r &= \left(\frac{3M_M}{4\pi\rho_m}\right)^{1/3}.
\end{align}$$
That looks surprisingly familiar...
Putting it together
We now plug our minimum $d$ and maximum $r$ into the visual angle equation
$$\begin{align} \alpha &= 2\arctan{\frac{2r}{2d}} \\
&=2\arctan{\frac{\left(\frac{3M_M}{4\pi\rho_m}\right)^{1/3}}{\left(\frac{9M_M}{4\pi\rho_m}\right)^{1/3}}} \\
&=2\arctan{(1/3^{1/3})} \\
&= 1.213 \text{ radians}.
\end{align}$$
The density terms conveniently cancel out, so we are left with a maximum value of 1.213 radians.
Conclusion
If there are two Earth-mass planets orbiting each other as binary planets, at the closest possible distance based on their Roche limits; they would appear to be 69 degrees of arc wide in the sky.
There are actually infinite solutions to this problem. A less dense Earth-mass 'moon' would be larger in radius, but would have to be farther away to not be pulled apart by gravity. So you can have some variation here, within the reasonable bounds of planetary density. For example, an Earth-mass moon at the distance of Luna would have to be 1320 kg/m$^3$ to be the maximum visual size. That is barely more than the density of water, so it is unlikely unless it is mostly of gas like a gas giant (gas midget?).
So this speaks to your gravitational effects problem. Since gravity falls off as the square of distance, a lower density object farther away would have minimal gravitational effects.
Also note that this minimum orbit is almost certainly not stable in the long run. Any reasonable moon would have to be smaller. But if artificial means are acceptable, then you can have a short term stable satellite take up nearly half of the sky.
