The Minkowski diagram captures time dilation even without invoking the Minkowski distance, provided we account for the different units/scales along the unprimed and the primed axes, see for example "Minkowski diagram in special relativity" (Wikipedia link). The calculation is exactly equivalent to that using the Minkowski distance as explained by WillO, but uses the Euclidian distance and usual trig/geometry.
Let the unit along the $x$ and $ct$ axes be $U$, and that along the $x'$ and $ct'$ axes be $U'$. They are related as
$$
U' = U \sqrt{\frac{1+\beta^2}{1-\beta^2}}
$$
for $\beta = v/c$ as usual.
In the case of the "pink" and "blue" events on your diagram, let the time difference between them in the unprimed frame be $ct$. This is just the Euclidian distance from "blue" to "pink" along the $ct$ axis in units of $U$. The time difference between these events as perceived in the primed frame is the Euclidian distance between the two intersections of green lines in units of $U'$. Let us first calculate this Euclidian distance according to usual geometry, in units of U.
In the triangle formed by "blue", "pink", and the 2nd intersection of green lines (first one being "blue" itself), denote $\overline{ct'}$ the desired Euclidian distance along the ct' axis and $\theta$ the angle between the $ct$ and $ct'$ axes. Calculate the other angles in terms of $\theta$ and apply the sine theorem to obtain
$$
\frac{ct}{\sin\left( \frac{\pi}{2}-2\theta \right)} = \frac{\overline{ct'}}{\sin\left(\frac{\pi}{2}+\theta \right)} \;\; \text{or} \;\; \frac{ct}{\cos 2\theta} = \frac{\overline{ct'}}{\cos \theta}
$$
Since $\tan \theta = \beta$, we have $\cos \theta = \frac{1}{\sqrt{1+\tan^2 \theta}} = \frac{1}{\sqrt{1+\beta^2}}$ and $\cos 2\theta = 2\cos^2 \theta -1 = \frac{1-\beta^2}{1+\beta^2}$, and we find
$$
\overline{ct'} = \frac{1}{\sqrt{1+\beta^2}} \frac{1+\beta^2}{1-\beta^2} ct = \frac{\sqrt{1+\beta^2}}{1-\beta^2} ct
$$
Now express $\overline{ct'}$ in units of $U'$ to obtain the correct time observed in the primed frame as
$$
ct' = \frac{\overline{ct'}\cdot U}{U'} = \frac{\sqrt{1+\beta^2}}{1-\beta^2} ct \sqrt{\frac{1-\beta^2}{1+\beta^2}} = \frac{ct}{\sqrt{1-\beta^2}}
$$
So, if the time difference between "blue" and "pink" in the unprimed frame is $ct$, the corresponding time difference $ct'$ in the primed frame is dilated by a factor $\gamma = \frac{1}{\sqrt{1-\beta^2}}$, which is the time dilation we were looking for.
In a similar way we can calculate the time dilation perceived in the unprimed frame relative to events in the primed frame.