In triangle $ABC$, $D$ is a point on $AB$ between $A$ and $B$, $E$ is a point on $AC$ between $A$ and $C$, and $F$ is a point on $BC$ between $B$ and $C$ such that $AF$, $BE$, and $CD$ all meet inside $\triangle ABC$ at a point $G$. Given that the area of $\triangle ABC$ is $15$, the area of $\triangle ABE$ is $5$, and the area of $\triangle ACD$ is $10$, compute the area of $\triangle ABF$.
1. Given the area of $\triangle ABC$ is $15$, the area of $\triangle ABE$ is $5$, and the area of $\triangle ACD$ is $10$.
2. We need to find the area of $\triangle ABF$.

First, let's analyze the given areas and their implications:
- The area of $\triangle ABE$ is $5$, so the area of $\triangle BCE$ is $15 - 5 = 10$.
- The area of $\triangle ACD$ is $10$, so the area of $\triangle BCD$ is $15 - 10 = 5$.

Next, we use the ratios of the areas to find the ratios of the segments:
- Since $\text{Area}[\triangle ABE] = 5$ and $\text{Area}[\triangle BCE] = 10$, we have:
  \[
  \frac{\text{Area}[\triangle ABE]}{\text{Area}[\triangle BCE]} = \frac{5}{10} = \frac{1}{2}
  \]
  This implies that $\frac{AE}{EC} = \frac{1}{2}$.

- Similarly, since $\text{Area}[\triangle ACD] = 10$ and $\text{Area}[\triangle BCD] = 5$, we have:
  \[
  \frac{\text{Area}[\triangle BCD]}{\text{Area}[\triangle ACD]} = \frac{5}{10} = \frac{1}{2}
  \]
  This implies that $\frac{BD}{DA} = \frac{1}{2}$.

Now, we apply Ceva's Theorem to the triangle $\triangle ABC$ with cevians $AF$, $BE$, and $CD$ intersecting at point $G$:
\[
\frac{AE}{EC} \cdot \frac{CF}{FB} \cdot \frac{BD}{DA} = 1
\]
Substituting the known ratios:
\[
\frac{1}{2} \cdot \frac{CF}{FB} \cdot \frac{1}{2} = 1
\]
Solving for $\frac{CF}{FB}$:
\[
\frac{1}{4} \cdot \frac{CF}{FB} = 1 \implies \frac{CF}{FB} = 4
\]

This implies that $\frac{FB}{CB} = \frac{FB}{CF + FB} = \frac{1}{4 + 1} = \frac{1}{5}$.

Finally, we calculate the area of $\triangle ABF$:
\[
\text{Area}[\triangle ABF] = \frac{FB}{CB} \cdot \text{Area}[\triangle ABC] = \frac{1}{5} \cdot 15 = 3
\]

The final answer is $\boxed{3}$.