Terry the Tiger lives on a cube-shaped world with edge length $2$. Thus he walks on the outer surface. He is tied, with a leash of length $2$, to a post located at the center of one of the faces of the cube. The surface area of the region that Terry can roam on the cube can be represented as $\frac{p \pi}{q} + a\sqrt{b}+c$ for integers $a, b, c, p, q$ where no integer square greater than $1$ divides $b, p$ and $q$ are coprime, and $q > 0$. What is $p + q + a + b + c$? (Terry can be at a location if the shortest distance along the surface of the cube between that point and the post is less than or equal to $2$.)

1. **Understanding the Problem:**
   Terry the Tiger is tied to a post at the center of one face of a cube with edge length 2. The leash length is also 2. We need to find the surface area of the region Terry can roam on the cube, expressed in the form \(\frac{p \pi}{q} + a\sqrt{b} + c\), and then determine \(p + q + a + b + c\).

2. **Visualizing the Problem:**
   Consider the net of the cube. The post is at the center of one face, and Terry can roam within a circle of radius 2 centered at this point. However, since the cube has finite boundaries, the circle will intersect the edges of the face and potentially extend onto adjacent faces.

3. **Calculating the Area on the Central Face:**
   The area of the circle on the central face is \(\pi \times 2^2 = 4\pi\). However, this circle extends beyond the edges of the face.

4. **Determining the Overlapping Regions:**
   The circle intersects the edges of the face, creating sectors on the adjacent faces. Each sector is a \(60^\circ\) sector of a circle with radius 2.

5. **Calculating the Area of the Sectors:**
   Each \(60^\circ\) sector has an area of:
   \[
   \text{Area of one sector} = \frac{60^\circ}{360^\circ} \times \pi \times 2^2 = \frac{1}{6} \times 4\pi = \frac{2\pi}{3}
   \]
   Since there are 4 such sectors, the total area of the sectors is:
   \[
   4 \times \frac{2\pi}{3} = \frac{8\pi}{3}
   \]

6. **Calculating the Area of the Triangular Regions:**
   Each triangular region is formed by an isosceles triangle with an isosceles right triangle cut out of its base. The angle \(\angle BOC\) is \(30^\circ\) because \(\triangle ABO\) is equilateral. The area of one such shape is:
   \[
   \text{Area of one shape} = \frac{2^2}{2} \cdot \sin(30^\circ) - \frac{1}{2} \cdot (\sqrt{3} - 1)^2 = 2 \cdot \frac{1}{2} - \frac{1}{2} \cdot (3 - 2\sqrt{3} + 1) = 1 - \frac{1}{2} \cdot (4 - 2\sqrt{3}) = 1 - 2 + \sqrt{3} = \sqrt{3} - 1
   \]
   Since there are 4 such shapes, the total area of these shapes is:
   \[
   4 \times (\sqrt{3} - 1) = 4\sqrt{3} - 4
   \]

7. **Summing Up the Areas:**
   The total area Terry can roam is the sum of the areas of the sectors and the triangular regions:
   \[
   \text{Total area} = \frac{8\pi}{3} + 4\sqrt{3} - 4
   \]

8. **Expressing in the Required Form:**
   The total area is already in the form \(\frac{p \pi}{q} + a\sqrt{b} + c\), where:
   \[
   p = 8, \quad q = 3, \quad a = 4, \quad b = 3, \quad c = -4
   \]

9. **Calculating \(p + q + a + b + c\):**
   \[
   p + q + a + b + c = 8 + 3 + 4 + 3 - 4 = 14
   \]

The final answer is \(\boxed{14}\)