Answer: $r = 1/3$.

Consider three circles $\omega_1$, $\omega_2$ and $\omega_3$ of radius $1$ with the centers $O_1$, $O_2$ and $O_3$ respectively and the equilateral triangle $ABC$ such that the side $AB$ touches the circles $\omega_1$ and $\omega_2$, the side $BC$ touches the circles $\omega_2$ and $\omega_3$, and the side $AC$ touches the circles $\omega_1$ and $\omega_3$. The part of the interior of the triangle $ABC$ which is not covered by circles is divided into seven parts of three types: the central part bounded by three circles, three side parts bounded by two circles and a side, and three corner parts bounded by a circle and two sides.

The central part can be completely covered with a side part, so it is senseless to enter the fourth circle there. Let's draw the circle of the largest radius $r$ that can be inscribed in the side part enclosed between the line $AB$ and circles $\omega_1$, $\omega_2$. Obviously, this circle touches $AB$, $\omega_1$ and $\omega_2$, so from the Pythagorean theorem we find $(1+r)^2 = (1-r)^2 + 1$, i.e. $r = 1/4$.

Let's draw the circle of the largest radius $r$ that can be inscribed in the corner part enclosed between $AC$, $CB$ and $\omega_3$. Obviously, this circle is tangent to $AC$, $CB$ and $\omega_3$, so its center $O$ lies on the bisector of angle $ACB$ and, since the leg opposite the angle at $30^\circ$ is half hypotenuse,
$$2 = CO_3 = CO + OO_3 = 2r + (r + 1),$$
whence $r = 1/3$. Therefore, in this case it is impossible to place a circle of radius greater than $1/3$ that doesn't intersect with the three already drawn. This means that for $r > 1/3$ it is impossible to say for sure that there will always be a circle that satisfies the condition.

At least one angle of the triangle $ABC$ is not greater than $60^\circ$, without loss of generality, let this angle be $ACB$. Let $\omega$ be that of the given three circles whose center is no farther from the vertex $C$ than the center of any of the remaining circles. If $\omega$ does not touch any side of the angle $ACB$, then replace it with a circle that: has the same center, is inside the angle $ACB$, and touches at least one of the sides of this angle (without loss of generality, let it touch the side $BC$ at the point $T$). If the resulting circle does not touch the side $AC$, then replace it with a circle that: touches the side $BC$ at the point $T$ and touches the side $AC$. Denote the resulting circle by $\Omega$, it is inside the corner $ACB$ and contains the circle $\omega$ inside itself. Consider the circle $\Gamma$ that touches the sides of the angle $ACB$ and the circle $\Omega$, and its center $O_1$ is closer to $C$ than the center $O$ of the circle $\Omega$. Denote by $r$ and $R$ the radii, and by $T_1$ and $T$ the touching points of $BC$ with the circles $\Gamma$ and $\Omega$. Then
$$R + r = OO_1 = OC - O_1C = \frac{R}{\sin \frac{\angle ABC}{2}} - \frac{r}{\sin \frac{\angle ABC}{2}} = \frac{R - r}{\sin \frac{\angle ABC}{2}}.$$
Consequently,
$$\frac{R - r}{R + r} = \sin \frac{\angle ABC}{2} \le \sin 30^\circ = \frac{1}{2}.$$
Solving this inequality we find that $r \ge R/3 \ge 1/3$. Hence circle of radius $1/3$ is guaranteed to be drawn inside the triangle $ABC$ so that it does not intersect with the three already drawn circles.