What is known is explained in C. Albanese, S. Lawi, Laplace transform of an integrated gometric Brownian motion, MPRF 11 (2005), 677-724, in particular in the paragraph of the Introduction beginning by A separate class of models...
Obviously, we can write
$$M_T = \mathbb{E} \left[ \left( \int_0^T 1 \, dW_t \right) \cdot \left( \int_0^T t \, dW_t \right) \right],$$
i.e. we are interested in the expectation of a product of stochastic integrals. Using the identity
$$a \cdot b = \frac{1}{4} ((a+b)^2-(a-b)^2) \tag{1}$$
for
$$a := \int_0^T 1 \, dW_t \qquad \quad b := \int_0^T t \, dW_t$$
we get
$$M_T = \frac{1}{4} \mathbb{E} \left[ \left(\int_0^T (1+t) \, dW_t \right)^2 \right] - \frac{1}{4} \mathbb{E} \left[ \left(\int_0^T (1-t) \, dW_t \right)^2 \right].$$
Applying Itô's isometry yields
$$\begin{align*} M_T &= \frac{1}{4} \mathbb{E} \left( \int_0^T (1+t)^2 \, dt \right) - \frac{1}{4} \mathbb{E} \left( \int_0^T (1-t)^2 \, dt \right) \\ &= \int_0^T \frac{1}{4} ((1+t)^2-(1-t)^2) \, dt \\ &\stackrel{(1)}{=} \int_0^T t \, dt. \end{align*}$$
In fact, we have shown the following (more general) statement:
Let $f,g \in L^2([0,T] \otimes \mathbb{P})$ be progressively measurable. Then $$\mathbb{E} \left[ \left( \int_0^T f(t) \, dW_t \right) \cdot \left( \int_0^T g(t) \, dW_t \right) \right] = \mathbb{E} \int_0^T g(t) \cdot f(t) \, dt.$$
Note that for $f=g$ this is the (standard version of) Itô's isometry.
Best Answer
Integration by parts looks like $$\int_a^b f(t)g'(W_t)dW_t = f(t)g(W_t)|_a^b - \int_a^b f'(t)g(W_t)dt - \frac{1}{2}\int_a^b f(t)g''(W_t)dt,$$
where $f(t)$ is a deterministic function and $g(W_t)$ is a function of a Wiener process. For the integral you want we can let $f(t) = 1$ and we get
$$\int_a^b g'(W_t)dW_t = g(W_t)|_a^b - \frac{1}{2}\int_a^b g''(W_t)dt.$$
So for the integral $\int_a^b W_tdW_t$ we can let $g(x) = \frac{x^{n+1}}{n+1}$ and this gives
$$\int_a^b W^n_tdW_t = \frac{W_t^{n+1}}{n+1}|_a^b - \frac{n}{2}\int_a^bW^{n-1}_t dt.$$