Inspired by Aniket Verma

We'll focus first on the square $S$ formed by the centers of the four circles in the upper left of the main square. The violet region is then that region of $S$ common to the four quarter-circles centered at each of the four corners of $S.$

Now divide this violet region into four identical portions with vertical and horizontal lines acting as axes of symmetry, (each of which pass through the center of the violet region).

Looking at top right portion, we see that it's area is found by first starting with a sector of the lower left circle that subtends an angle of $30^{\circ},$ and then subtracting the areas of two congruent triangles each with a base length of $4$ and a height of $2\sqrt{2}\sin(15^{\circ}) = \sqrt{3} - 1.$ This portion then has an area of

$\dfrac{1}{2}*4^{2}*\dfrac{\pi}{6} - 2*\dfrac{1}{2}*4*(\sqrt{3} - 1) = \dfrac{4\pi}{3} - 4\sqrt{3} + 4.$

Now the four "whole" violet regions will then be composed of $16$ of these portions, and hence the desired area value is

$64*(\dfrac{\pi}{3} - \sqrt{3} + 1) = \boxed{20.169}$ to 3 decimal places.

Each purple part could be subdivided into a square and 4 arc slivers extending from each side. Let $A_1$ be the surface area of the square and $A_2$ that of the arcs. The total area of all four purple shapes will be: $A_0 = 4A_1 + 16A_2$

$\underline{1- Square}$

To calculate the area of the square, we first calculate the side, segment $AB$ in the image at left.

Law of cosines: $AB^2 = OA^2 + OB^2 - 2.OA.OB.cos(30^\circ)$

$AB^2 = r^2 + r^2 - 2.r^2.\sqrt {3}/2 \Rightarrow AB = r \sqrt{2-\sqrt{3}}$

$\boxed{A_1 = r^2(2-\sqrt{3})}$

$\underline{2- Arcs}$

The surface area of a blue arc is the surface of the circular sector, $A_{21}$ minus the surface of triangle $OAB$ , $A_{22}$

$A_{21} = \frac{\pi r^2}{12}$

$A_{22} = OH.AB/2$

$\cos 15^ \circ = OH/OA \Rightarrow OH=r(\sqrt{6} + \sqrt{2})/4$

$A_{22} = r^2 (\sqrt{2-\sqrt{3}})(\sqrt{6} + \sqrt{2})/8 = r^2/4$ (It's a fun little factoring exercise)

$A_{2} = A_{21} - A_{22} = \frac{\pi r^2}{12} - \frac{r^2}{4} \Rightarrow \boxed{A_2 = r^2 (\frac{\pi}{12} - \frac{1}{4})}$

$\underline{Total}$

$A_0 = 4A_1 + 16A_2 = 4r^2(2-\sqrt{3}) + 16r^2 (\frac{\pi}{12} - \frac{1}{4}) \Rightarrow \boxed{A_0 = 4r^2(1- \sqrt{3} + \frac{\pi}{3})}$

For $r=4, A_0 = 20.169$

I use calculus to solve this. Please refer to picture for reference. I first calculate area A by integrating y=4-sqrt(8x-x^2) from 2 to 4. This function is the circle to the top right. A= 0.34711.

Next, calculate area B. This is just simply the quarter circle minus half of the 4x4 square. Therefore, area B= 1/4 pi 16-8=4.566370

Next, calculate area C. It is 8 -Area B= 8=4.566370=3.43363

Now, we can calculate D. D= C- 4*Area A= 3.43363- 4(0.34711)=2.04519

Hence, E( violet area) = 4* (2 AreaB) -2 (AreaD)= 4*(9.13274-4.09038)=20.16944

A quarter of one of the four regions is confined by x² + y² = 16; y = 2 and x = 2. This yields to dA = ( sqrt(16 - x²) - 2)dx. Integrating from x = 2 tot x = sqrt(12) yields to the surface of this area. Multiply by 16 to get the answer.