I don't know whether this math will be clarifying forr you, but it might help you understand why "It is strongly influenced by the most abundant species." It is all about finding the best p's that does the minimization for $D$. You can skip to conclusion if you feel like skipping all the math.
You want to find the critical point (where the function attains its minimum and maximum) of:
$$D = p_{1}^2+p_{2}^2+...+p_{n}^2$$
given ${p_{1}+p_{2}+...+p_{n}=1}$, for $p_{i}$ of species.
Lagrange Multiplier method is one of the techniques to find the values given to $p_{1}, p_{2}, ...p_{n}$ to make $p_{1}^2+p_{2}^2+...+p_{n}^2$ go to maximum or minimum given some constraint on this.
Assume:
$$f = p_{1}^2+p_{2}^2+...+p_{n}^2$$
$$g = p_{1}+p_{2}+...p_{n}$$
According to Lagrange multiplier:
$$\partial_{p_{1},..p_{n}} f = \lambda \partial_{p_{1},..p_{n}} g$$
We have to find the value of $\lambda$.
So after partial fractions step, we will end up with these equations:
$$np_{1} = \lambda ; np_{2} = \lambda; ... np_{n} = \lambda$$
and we already know that $p_{1}+p_{2}+...+p_{n} = 1$
From above, we know that:
$$p_{1} = p_{2} = ... = p_{n} = \lambda/n$$ we get $\lambda = 2/n$ and from this the minimum values of the function:
$$f = D = p_{1}^2+p_{2}^2+...+p_{n}^2$$
is when all $p_{i} = 1/n $ for i is denoting each species in ${1,...,n}$
Conclusion:
So $D$ value attains its minimum when all terms $p_{i}$ are equal. As any value goes away from $1/n$ ('n' is the total number of individuals of species you got)the value increases drastically as it is being squared in $D$.
Sample example will be like $0.5^2 + 0.5^2 = 0.25$ but a change from this say $0.6,0.4$ increases $0.5$ term by $0.11$ but decreases $0.4$ term by only $0.09$.
Hope this helps.
