Using the Ito Formula
The general approach that often works for these kinds of question is to search for functions such that their Ito differential contains the terms that we are interested in. In your case, we are looking for a function $f(t, x)$ such that $f_t(t, x) = t x$. Let
\begin{equation}
f(t, x) = \frac{1}{2} t^2 x
\end{equation}
with
\begin{equation}
f_t(t, x) = t x, \qquad f_x(t, x) = \frac{1}{2} t^2, \qquad f_{xx}(t, x) = 0.
\end{equation}
Applying the Ito formula yields
\begin{equation}
\frac{1}{2} T^2 W_T = \int_0^T u W_u \mathrm{d}u + \frac{1}{2} \int_0^T u^2 \mathrm{d}W_u
\end{equation}
or
\begin{equation}
\int_0^T u W_u \mathrm{d}u = \frac{1}{2} \int_0^T \left( T^2 - u^2 \right) \mathrm{d}W_u.
\end{equation}
The Ito integral of a deterministic integrand is normally distributed with zero mean and variance
\begin{equation}
\text{Var} \left( \int_0^T (T^2 - u^2) \mathrm{d}W_u \right) = \int_0^T \left( T^2 - u^2 \right)^2 \mathrm{d}u = \frac{8}{15} T^5
\end{equation}
and we conclude that
\begin{equation}
\int_0^T u W_u \mathrm{d}u \sim \mathcal{N} \left( 0, \frac{2}{15} T^5 \right).
\end{equation}
Discretizing the Integral
Alternatively, we could write
\begin{equation}
\int_0^T u W_u \mathrm{d}u = \lim_{n \rightarrow \infty} X_n, \qquad X_n = \sum_{i = 1}^n \underbrace{t_i W_{t_i}}_{Y_i} \Delta_n,
\end{equation}
where $\Delta_n = T / n$ and $t_i = i \Delta_n$. Then, each $Y_i$ is normal with mean zero and covariance
\begin{equation}
\text{Cov} \left( Y_i, Y_j \right) = t_i t_j \min \left\{ t_i, t_j \right\} = i j \min \{ i, j \} \Delta_n^3 .
\end{equation}
Letting $\bar{Y}_n$ be the corresponding column-vector with elements $\left( Y_1, Y_2, \ldots, Y_n \right)'$, we get in matrix form
\begin{equation}
\bar{\Sigma}_n = \mathbb{E} \left[ \bar{Y}_n \bar{Y}_n' \right] = \Delta_n^3 \left[ \begin{array}{c c c c} 1 & 2 & \dots & n\\ 2 & 8 & \dots & 4 n\\ 3 & 12 & \ddots & \vdots\\ n & 4 n & \dots & n^3 \end{array} \right]
\end{equation}
As the weighted sum of normally distributed random variables is itself normally distributed, it follows that $X_n \sim \mathcal{N} \left( 0, \Delta_n \bar{1}_n \bar{\Sigma}_n \bar{1}_n' \Delta_n \right)$, where $\bar{1}_n$ is an $n$-dimensional column vector of ones. We have
\begin{eqnarray}
\text{Var} \left( X_n \right) & = & \Delta_n^5 \sum_{i = 1}^n \left( i \sum_{j = 1}^i j^2 + i^2 \sum_{j = i + 1}^n j \right)\\
& = & \Delta_n^5 \left( \sum_{i = 1}^n \left( \frac{1}{3} i^4 + \mathcal{O} \left( i^3 \right) + i^2 \left( \frac{1}{2} n^2 - \frac{1}{2} i^2 + \mathcal{O}(i) + \mathcal{O}(n) \right) \right) \right)\\
& = & \Delta_n^5 \left( \frac{1}{15} n^5 + \frac{1}{6} n^5 - \frac{1}{10} n^5 + \mathcal{O}\left( n^4 \right) \right) \\
& = & \Delta_n^5 \left( \frac{2}{15} n^5 + \mathcal{O}\left( n^4 \right) \right)
\end{eqnarray}
Note that $\Delta_n^5 = \mathcal{O} \left( n^{-5} \right)$. Thus, we have
\begin{equation}
\lim_{n \rightarrow \infty} \text{Var} \left( X_n \right) = \frac{2}{15} T^5,
\end{equation}
just like before.