The Turán graph $T(n,r)$ is an $r$-partite graph whose parts are as nearly equal in size as possible. Let $n=pr+s$, where $p$ and $s$ are integers, and $0\le s<r$; then $T(n,r)$ has $s$ parts of size $p+1$ and $r-s$ parts of size $p$. Thus, $s\binom{p+1}2+(r-s)\binom{p}2$ of the $\binom{n}2$ pairs of vertices of $T(n,r)$ are within a single part and are not connected by an edge, and the number of edges of $T(n,r)$ is therefore
$$\begin{align*}
\binom{n}2-s\binom{p+1}2-(r-s)\binom{p}2&=\frac{n(n-1)-sp(p+1)-(r-s)p(p-1)}2\\\\
&=\frac{n^2-s-2ps-p^2r}2\\\\
&=\frac{p^2r^2+2prs+s^2-p^2r-2ps-s}2\\\\
&=\frac{(r-1)(p^2r+2ps)}2+\binom{s}2\\\\
&=\frac{(r-1)(p^2r^2+2prs)}{2r}+\binom{s}2\\\\
&=\frac{(r-1)(n^2-s^2)}{2r}+\binom{s}2\;.\tag{1}
\end{align*}$$
Contrary to the assertions in Wikipedia and MathWorld, this is not necessarily equal to
$$\left\lfloor\frac{(r-1)n^2}{2r}\right\rfloor\;,\tag{2}$$
as may be seen by considering the case $n=12,r=8$. $T(12,8)$ evidently has $4$ vertices of degree $11$ and $8$ of degree $10$, for a total of $\frac12(44+80)=62$ edges, which agrees with $(1)$, while $(2)$ evaluates to $63$. It is true, however, that
$$\frac{(r-1)(n^2-s^2)}{2r}+\binom{s}2<\frac{(r-1)n^2}{2r}$$
and hence that
$$\frac{(r-1)(n^2-s^2)}{2r}+\binom{s}2\le\left\lfloor\frac{(r-1)n^2}{2r}\right\rfloor\;,$$
since
$$\binom{s}2=\frac{s^2-s}2=\frac{rs^2-rs}{2r}<\frac{(r-1)s^2}{2r}\;.$$
A lower bound on the number of edges in $T(n,r)$ is
$$\frac{(r-1)n^2}{2r}-\frac{n}4\;.$$
Best Answer
I recall that the number $|E(T^r(n))|$ can be easily calculated exactly. Let $n=qr+s$, where $0\le s<r$. Then the $r$-partition of the vertex set of the graph $T^r(n)$ has $s$ parts of size $q+1$ and $r-s$ parts of size $q$. Thus
$$|E(T^r(n))|={s \choose 2}(q+1)^2+s(r-s)q(q+1)+ {r-s \choose 2}q^2=\frac 12\left( q^2r^2-q^2r+2sqr-2sq+s^2-s\right)=\frac 12\left(n^2-q(n+s)-s\right)= \frac 12\left(n^2-\frac {(n-s)(n+s)}{r}-s\right)= \frac 12\left(n^2-\frac {(n^2-s^2)}{r}-s\right).$$
Thus $$\left|E(T^r(n))|-\frac{n^2}{2}\left(1-\frac 1r\right)\right|=\frac{s(r-s)}{2r}\le \left(\frac {s+r-s}2\right)^2\frac 1{2r}=\frac r8.$$