1. Identify the position of the circle centers:
The centers of the circles $C_1, C_2, \ldots$ must lie on the internal bisector $AM$ of $\angle BAC$, since their distances from the sides $AB$ and $AC$ are equal.

2. Establish the relationship between the radii:
Consider the first two circles $C_1$ and $C_2$. Let the radius of $C_1$ be $x$ and the radius of $C_2$ be $y$. Since $C_2$ is externally tangent to $C_1$ and both are tangent to $AB$ and $AC$, the centers of these circles form a right triangle with the internal bisector $AM$. The hypotenuse of this triangle is $x + y$ and the leg opposite the $30^\circ$ angle is $y - x$. Using the ratio for a $30^\circ-60^\circ-90^\circ$ triangle, we have:
$$\frac{y - x}{x + y} = \frac{1}{2}$$
Solving for $y$, we get:
$$2(y - x) = x + y \implies 2y - 2x = x + y \implies 2y - y = x + 2x \implies y = 3x$$

3. Iterate the relationship for subsequent circles:
For $n = 2, 3, \ldots$, the radius of the $n$-th circle $C_n$ can be expressed as $r_n = 3^{n-1}x$. The distance from $A$ to the center of the $n$-th circle $O_n$ is the sum of the radii of all previous circles plus the radius of the $n$-th circle:
$$\overline{AP_n} = x + (x + 3x) + (3x + 9x) + \cdots + (3^{n-2}x + 3^{n-1}x) + 3^{n-1}x$$
Simplifying, we get:
$$\overline{AP_n} = x \left(1 + 4 + 12 + \cdots + 3^{n-1} + 3^{n-1}\right)$$
This is a geometric series with the first term $a = 1$ and common ratio $r = 3$:
$$\overline{AP_n} = x \left(1 + 4 + 12 + \cdots + 3^{n-1} + 3^{n-1}\right) = x \left(1 + 3 \sum_{k=0}^{n-2} 3^k\right)$$
The sum of the geometric series is:
$$\sum_{k=0}^{n-2} 3^k = \frac{3^{n-1} - 1}{2}$$
Therefore:
$$\overline{AP_n} = x \left(1 + 3 \cdot \frac{3^{n-1} - 1}{2}\right) = x \left(1 + \frac{3^{n} - 3}{2}\right) = x \left(\frac{3^n + 1}{2}\right)$$

4. Determine the condition for 4 circles to fit inside the triangle:
The length of the internal bisector $AM$ of $\angle BAC$ in an equilateral triangle with side length 3 is:
$$AM = \frac{\sqrt{3}}{2} \cdot 3 = \frac{3\sqrt{3}}{2}$$
We need:
$$\overline{AP_4} < \overline{AM} < \overline{AP_5}$$
Substituting the expressions for $\overline{AP_4}$ and $\overline{AP_5}$:
$$x \left(\frac{3^4 + 1}{2}\right) < \frac{3\sqrt{3}}{2} < x \left(\frac{3^5 + 1}{2}\right)$$
Simplifying:
$$x \left(\frac{81 + 1}{2}\right) < \frac{3\sqrt{3}}{2} < x \left(\frac{243 + 1}{2}\right)$$
$$x \cdot 41 < \frac{3\sqrt{3}}{2} < x \cdot 122$$
Solving for $x$:
$$x < \frac{3\sqrt{3}}{82} \quad \text{and} \quad x > \frac{3\sqrt{3}}{244}$$
Therefore, the possible values for $x$ are:
$$\frac{3\sqrt{3}}{244} < x < \frac{3\sqrt{3}}{82}$$

The final answer is $\boxed{\frac{3\sqrt{3}}{244} < x < \frac{3\sqrt{3}}{82}}$.