For convenience let the interval be $[0,1]$, and let's look at
the case $f(t) = t^p$.  We have $E_n(t^p) = 0$ for $p \le 1$, while by Faulhaber's formula we have 
$$\eqalign{E_n(t^p) &= \dfrac{1}{p+1} + \dfrac{1}{2n} - \dfrac{1}{n^{p+1}} \sum_{i=1}^{n} i^p\cr
&= - \sum_{j=0}^{p-2} {p \choose j} \dfrac{B_{p-j}}{j+1} n^{j-p} }$$
where $B_{p-j}$ is a Bernoulli number.
For any polynomial $f$, $E_n(f)$ is a linear combination of these, and for its asymptotic behaviour as $n \to \infty$ we take the least negative power of $n$ whose coefficient does not cancel out: $E_n(f) \sim c n^{-k}$ for some nonzero constant $c$ and some integer $k \ge 2$.  In particular its absolute value is monotonically decreasing for sufficiently large $n$.

EDIT: For smooth functions the behaviour can be more complicated.  Consider e.g. 
a function $f$ that is periodic with period $1$.  If we write $f$ in its
Fourier series
$$ f(x) = \sum_{m=-\infty}^\infty c_m \exp(2\pi i m x)$$
and note that $T_n(\exp(2\pi i m x) = 0$ unless $m$ is a multiple of $n$, in which case $T_n(\exp(2\pi i m x) = 1$.  Thus
$E_n(f) = - \sum_{k \ne 0} c_{kn}$, which need not be monotonic.