How to compute the solid angle of the tangent cone of an intersection of surfaces? We can start with a relatively simple question in $\mathbb{R}^2$: if two curves $\gamma_1, \gamma_2$ intersect at some point $p$ what is the "angle" they intersect at?
Naturally one can look at the point $p = (p_x , y_x)$ and calculate $\gamma_1'(p_x)$ to get the slope of the first curve (and similarly for the second) and then compute
$$ \left| \tan^{-1} (\gamma_1'(p_x)) - \tan^{-1}(\gamma_2'(p_x)) \right|$$
To actually calculate the "angle".
Now angles in $\mathbb{R}^3$ are considerably more complex beasts, they are a real number ranging from $(0, 2\pi)$ (although their natural domain is the area of the unit sphere which is $4\pi$ like the natural domain of angles in $\mathbb{R}^2$ is the circumference of the unit circle which is $2\pi$).
So the question...
is suppose we have 3 surfaces $\gamma_1, \gamma_2, \gamma_3$ intersecting at a point $p = (p_x, p_y, p_z)$ how can we calculate the "solid angle" of the natural cone at their intersection point?
This question gets hard because it is NOT ENOUGH to look at tangent planes, when forming these cones, whereas in $\mathbb{R}^2$ it was enough to just look at tangent lines. We actually have to do some work to FIND the tangent cone boundary shape.
An approach:
If you had a specific example of 3 such planes you could try draw the image out, and draw a plane "orthogonal" to the cone (how to define that abstractly/generically?) and then take the limit as the plane approaches the intersection point of the ratio of the image in the plane intersection divided by distance to the intersection point to come up with the correctly scaled "shape" of the cone tangent boundary and then try to intersect this shape with the unit sphere (taking a surface integral for the area) but actually carrying this out in calculation is difficult and it seems like theres a lot of implicit assumptions here so writing such a formula in generality feels difficult.
 A: Locally under high magnification each surface looks like a plane that splits space into two signed halves (+ or -).
Once you have chosen a sign for each surface , there is a well-defined region of space to consider that is the intersection of three half-spaces ( a cone in space  with 3 walls)
The solid angle  sought is proportional to  the surface area of the region R on the  sphere  that lies inside this prism. On the sphere, each wall cuts the sphere in a great circle. So your spherical region R is a geodesic triangle on the sphere. There is a formula due to Girard that allows you to find this surface area easily.
A: Instead of our 'tan' inverse formula we could have solved the 2d problem another way.
We could instead consider the circle of radius $r$ centered at the intersection point and evaluating
$$ \int_{\text{arc between $\left( \gamma_1 \cup \gamma_2 \right) \cap (x-p_x)^2 + (y-p_y)^2 = r^2 $  }} 1  $$
Which measures the length of the circular arc constrained between $\gamma_1, \gamma_2$.
We could take then consider
$$ \lim_{r \rightarrow 0} \int_{\text{arc between $\left( \gamma_1 \cup \gamma_2 \right) \cap (x-p_x)^2 + (y-p_y)^2 = r^2 $  }} 1  $$
But this would go to 0, to get the angle we need to divide by $r$ along the way to rescale our units giving us that our "angle" operator at the point is:
$$ \lim_{r \rightarrow 0} \frac{\int_{\text{arc between $\left( \gamma_1 \cup \gamma_2 \right) \cap (x-p_x)^2 + (y-p_y)^2 = r^2 $  }} 1}{r}   $$
Now in the 3-dimensional world the solid angle is proportional to $r^2$ instead of $r$ so the formula we are targeting is:
$$ \lim_{r \rightarrow 0} \frac{\int_{\text{area between $ \left( \gamma_1 \cup \gamma_2  \cup \gamma_3 \right) \cap (x-p_x)^2 + (y-p_y)^2 + (z - p_z)^2 = r^2 $  }} 1}{r^2}   $$
In the event that these surfaces locally are flat around the point this recovers @MathWonk's answer.
This is still messy so I will try to clean it up
