You are right to think of the probabilities as areas, but the set of points closer to the center is not a triangle. It's actually a weird shape with three curved edges, and the curves are parabolas.
The set of points equidistant from a line $D$ and a fixed point $F$ is a parabola. The point $F$ is called the focus of the parabola, and the line $D$ is called the directrix. You can read more about that here.
In your problem, if we think of the center of the triangle $T$ as the focus, then we can extend each of the three edges to give three lines that correspond to the directrices of three parabolas.
Any point inside the area enclosed by the three parabolas will be closer to the center of $T$ than to any of the edges of $T$. The answer to your question is therefore the area enclosed by the three parabolas, divided by the area of the triangle.
Let's call $F$ the center of $T$. Let $A$, $B$, $C$, $D$, $G$, and $H$ be points as labeled in this diagram:
The probability you're looking for is the same as the probability that a point chosen at random from $\triangle CFD$ is closer to $F$ than to edge $CD$. The green parabola is the set of points that are the same distance to $F$ as to edge $CD$.
Without loss of generality, we may assume that point $C$ is the origin $(0,0)$ and that the triangle has side length $1$. Let $f(x)$ be equation describing the parabola in green.
By similarity, we see that $$\overline{CG}=\overline{GH}=\overline{HD}=1/3$$
An equilateral triangle with side length $1$ has area $\sqrt{3}/4$, so that means $\triangle CFD$ has area $\sqrt{3}/12$. The sum of the areas of $\triangle CAG$ and $\triangle DBH$ must be four ninths of that, or $\sqrt{3}/27$.
$$P\left(\text{point is closer to center}\right) = \displaystyle\frac{\frac{\sqrt{3}}{12} - \frac{\sqrt{3}}{27} - \displaystyle\int_{1/3}^{2/3} f(x) \,\mathrm{d}x}{\sqrt{3}/12}$$
We know three points that the parabola $f(x)$ passes through. This lets us create a system of equations with three variables (the coefficients of $f(x)$) and three equations. This gives
$$f(x) = \sqrt{3}x^2 - \sqrt{3}x + \frac{\sqrt{3}}{3}$$
The integral of this function from $1/3$ to $2/3$ is $$\int_{1/3}^{2/3} \left(\sqrt{3}x^2 - \sqrt{3}x + \frac{\sqrt{3}}{3}\right) \,\mathrm{d}x = \frac{5}{54\sqrt{3}}$$
This gives our final answer of $$P\left(\text{point is closer to center}\right) = \boxed{\frac{5}{27}}$$
If we set (see your first figure) $BG=x$, $CG=3-x$, $\angle ABC=\angle ACB=\alpha$, $\angle CGI=\theta$, $GI=IH=HG=l$, then by the sine rule applied to triangles $GCI$ and $BGH$ we obtain:
$$
{l\over\sin\alpha}={3-x\over\sin(\alpha+\theta)}={x\over\sin(60°+\theta-\alpha)}.
$$
We can solve for $x$ to find:
$$
x={3\over2}\left(1-{\tan(\alpha-30°)\over\tan(30°+\theta)} \right).
$$
Substituting that into the expression for $l$ and differentiating one gets
$$
{dl\over d\theta}=-{3\over2}{\sin\alpha\over\sin(60°+\theta-\alpha)}
\big(1-\tan(\alpha-30°)\big)\cot(30°+\theta).
$$
This vanishes only for $\theta=60°$, but that is a minimum (as one can deduce from the sign of $dl/d\theta$). Hence maxima are attained at the boundary values of $\theta$, which are the values for which $x=0$ or $x=3$, that is $\theta=\alpha-60°$ and $\theta=180°-\alpha$.
This gives as a maximum for $l$:
$$
l_\max={3\sin\alpha\over\sin(2\alpha-60°)}.
$$
Best Answer
It turns out that it is possible to place eight points in an equilateral triangle of side $1$ so that the minimum distance is greater than $\frac13$. The diagram below shows how the points can be arranged.
Points $A$, $B$ and $C$ are the vertices of the equilateral triangle. Point $D$ is the midpoint of $AB$. Points $E$ and $F$ are on sides $AC$ and $BC$, respectively, each at a distance $d$ (to be determined) from $C$.
Points $G$ and $H$ are the circumcenters of triangles $ADE$ and $BDF$, respectively. The value of $d$ will be chosen so that the circumradius of each circle is also $d$, which will then be the minimum distance between any pair of the eight points.
It remains to compute $d$. Since $\angle DAE = \frac{\pi}{3}$, the central angle $\angle DGE$ subtending the same arc is $\frac{2\pi}{3}$. $GD = GE = d$, so an application of the law of cosines to $\angle DGE$ finds that $DE = \sqrt{3}\,d$. $AE = 1 - d$ and $AD = \frac12$, so using the law of cosines at $\angle DAE$ gives $$ (1-d)^2 + (\frac12)^2 - 2(1-d)(\frac12)\cos\frac{\pi}{3} = (\sqrt{3}\,d)^2, $$ which simplifies to $8d^2 + 6d - 3 = 0$. The positive root is $d = \dfrac{\sqrt{33}-3}{8} \approx 0.34307$. Since this is greater than $\frac13$, this arrangement satisfies the claimed condition.
(I first found a different arrangement that worked, although with a smaller minimum distance. It then occurred to me to checking the best-known packings of circles into an equilateral triangle at Erich Friedman's Packing Center. The packing for eight circles, when adapted to points, was both better and easier to explain than the one I had found earlier.)