The key point is to write $B_T^2$ in a clever way:
$$\begin{align*} B_T^2 &= \sum_{j=1}^n (B_{t_j}^2-B_{t_{j-1}}^2) \\ &= \sum_{j=1}^n \big( (B_{t_j}-B_{t_j^{\ast}})+B_{t_j^{\ast}} \big)^2 - \big( (B_{t_{j-1}}-B_{t_j^{\ast}})+B_{t_j^{\ast}} \big)^2 \\ &= 2 \sum_{j=1}^n B_{t_j^{\ast}} (B_{t_j}-B_{t_{j-1}}) + \sum_{j=1}^n (B_{t_j}-B_{t_j^{\ast}})^2- \sum_{j=1}^n (B_{t_{j-1}}-B_{t_j^{\ast}})^2 \tag{1} \end{align*}$$
where $t_j^{\ast}:= \frac{1}{2}(t_j+t_{j-1})$. Now, by definition, the first term on the right-hand side converges to the Stratonovich integral
$$\int_0^T B_t \circ dB_t.$$
Consequently, it remains to show that the remaining terms converge to $0$ as the mesh size $|\Pi|$ converges to $0$. To this end, we set
$$S_n := \sum_{j=1}^n (B_{t_j}-B_{t_j^{\ast}})^2$$
and note that by the stationarity of the increments
$$\mathbb{E}(S_n) = \mathbb{E} \left( \sum_{j=1}^n (B_{t_j}-B_{t_j^{\ast}})^2 \right) = \sum_{j=1}^n (t_j-t_j^{\ast}) = \frac{T}{2}.$$
Consequently, by the independence of the increments,
$$\begin{align*} \mathbb{E} \left[ \left( \sum_{j=1}^n (B_{t_j}-B_{t_j^{\ast}})^2 - \frac{T}{2} \right)^2 \right] &= \text{Var} \, (S_n) \\ &= \sum_{j=1}^n \text{Var} \, \left[ (B_{t_j}-B_{t_j^{\ast}})^2 - (t_j-t_j^{\ast}) \right]. \end{align*}$$
Using the scaling property, i.e. $B_t \sim \sqrt{t} B_1$ for any $t \geq 0$, it is therefore not difficult to see that this expression converges to $0$ as $|\Pi| \to 0$. In fact, we find that
$$\mathbb{E} \left[ \left( \sum_{j=1}^n (B_{t_j}-B_{t_j^{\ast}})^2 - \frac{T}{2} \right)^2 \right] \leq C|\Pi| T$$
for $C:= \mathbb{E}((B_1^2-1)^2)<\infty$. Hence,
$$\sum_{j=1}^n (B_{t_j}-B_{t_j^{\ast}})^2 \stackrel{L^2}{\to} \frac{T}{2}.$$
Exactly the same calculation goes through for the third addend in $(1)$. Finally, we conclude
$$B_T^2 = 2\int_0^T B_t \circ dB_t + \frac{T}{2} - \frac{T}{2} = 2\int_0^T B_t \circ dB_t.$$
Best Answer
Your expression for Stratanovich integral is wrong - the correct sum should be
$$\sum\frac{W(t_{j+1})+W(t_j)}{2}(W(t_{j+1})-W(t_j))=\frac12\sum W^2(t_{j+1})-W^2(t_j)=\frac12 W^2(T)-\frac12 W^2(0).$$