Hint. You may obtain an explicit formula for $T_n$, by observing that
$$
(T_n+1)=2(T_{n-1}+1),\quad n=1,2,3,\cdots.
$$
Then deducing an asymptotic expansion is straightforward.
I think your Level $3$ is wrong: it should derive from the Level $2$ terms $2T(n/2)$ and $6T(n/3)$, not $T(n/2)$ and $T(n/3)$. So, combining the terms within each level, we have:
\begin{align}
\text{Level } 1: & \qquad \qquad n^2 \\
\text{Level } 2: & \qquad \left(\frac{7}{6}\right) n^2 \\
\text{Level } 3: & \qquad \left(\frac{7}{6}\right)^2 n^2 \\
\cdots & \qquad \cdots \\
\text{Level } k: & \qquad \left(\frac{7}{6}\right)^{k-1} n^2 \\
\end{align}
A lower bound is found via the "shortest path" through the levels from $n$ to $1$:
$$n,\; n\left(\dfrac{1}{3}\right),\; n\left(\dfrac{1}{3}\right)^2,\; n\left(\dfrac{1}{3}\right)^3,\;\ldots,\; 1 = \; n\left(\dfrac{1}{3}\right)^{\log_3 n}.$$
This gives a lower asymptotic bound:
\begin{align}
T(n) &\geq n^2\left( 1 + \left(\frac{7}{6}\right) + \left(\frac{7}{6}\right)^{2} + \cdots + \left(\frac{7}{6}\right)^{\log_3 n} \right) \\
&= n^2 \dfrac{\left(7/6\right)^{1+\log_3 n} - 1}{(7/6) - 1} \\
&= 7n^2 \left(7/6\right)^{\log_3 n} - 6n^2 \\
&= 7n^{2+\log_3 \left(7/6\right)} - 6n^2 \qquad\text{using $x^{\log_a y} = y^{\log_a x}$} \\
& \\
\therefore\quad T(n) &= \Omega\left( n^{2+\log_3 \left(7/6\right)} \right) \qquad\text{since $\log_3 (7/6) \gt 0$}.
\end{align}
An upper bound is found via the "longest path" through the levels from $n$ to $2$ (not to $1$ since the relation is valid for $n\geq 3$):
$$n,\; n\left(\dfrac{1}{2}\right),\; n\left(\dfrac{1}{2}\right)^2,\; n\left(\dfrac{1}{2}\right)^3,\;\ldots,\; 2 = \; n\left(\dfrac{1}{2}\right)^{\log_2 n \;-\; 1}.$$
This gives an upper asymptotic bound:
\begin{align}
T(n) &\leq n^2\left( 1 + \left(\frac{7}{6}\right) + \left(\frac{7}{6}\right)^{2} + \cdots + \left(\frac{7}{6}\right)^{\log_2 n \;-\; 1} \right) \\
&= n^2 \dfrac{\left(7/6\right)^{\log_2 n} - 1}{(7/6) - 1} \\
&= 6n^2 \left(7/6\right)^{\log_2 n} - 6n^2 \\
&= 6n^{2+\log_2 \left(7/6\right)} - 6n^2 \qquad\text{using $x^{\log_a y} = y^{\log_a x}$} \\
& \\
\therefore\quad T(n) &= O\left( n^{2+\log_2 \left(7/6\right)} \right) \qquad\text{since $\log_2 (7/6) \gt 0$}.
\end{align}
These two bounds can be verified inductively: for some constant $c$,
\begin{align}
T(n) & \geq 2c\left( \dfrac{n}{2}\right)^{2+\log_3(7/6)} + 6c\left( \dfrac{n}{3}\right)^{2+\log_3(7/6)} \\
&= cn^{2+\log_3(7/6)} \left[ \dfrac{1}{2}\cdot 2^{\log_3 (6/7)} + \dfrac{4}{7} \right] \\
&\geq cn^{2+\log_3(7/6)} \qquad\text{since $\left[ \dfrac{1}{2}\cdot 2^{\log_3 (6/7)} + \dfrac{4}{7} \right] \approx 1.025$} \\
& \\
\text{and:} \\
& \\
T(n) & \leq 2c\left( \dfrac{n}{2}\right)^{2+\log_2(7/6)} + 6c\left( \dfrac{n}{3}\right)^{2+\log_2(7/6)} + n^2 \\
&= n^{2+\log_2(7/6)} \left[ c\left(\dfrac{3}{7} + \dfrac{2}{3}\cdot 3^{\log_2 (6/7)} \right) + n^{\log_2(6/7)} \right] \\
&\geq cn^{2+\log_2(7/6)} \qquad\text{for $c\left(\dfrac{3}{7} + \dfrac{2}{3}\cdot 3^{\log_2 (6/7)} \right) + n^{\log_2(6/7)} \leq c$} \\
& \qquad\qquad\qquad\qquad\qquad\approx 0.95c + \dfrac{1}{n^{0.22}} \leq c \implies c\geq \dfrac{20}{n^{0.22}} \\
& \qquad\qquad\qquad\qquad\text{so $c = 20$ is sufficient.}
\end{align}
Best Answer
$\Theta(n2^n)$ can’t be correct. Let’s take a look at what happens without the asymptotics, say with $T(n)=2T(n-1)+n$ and $T(0)=1$. Then
$$\begin{align*} T(n)&=2T(n-1)+n\\ &=4T(n-2)+2(n-1)+n\\ &=8T(n-3)+4(n-2)+2(n-1)+n\\ &\,\vdots\\ &=2^kT(n-k)+\sum_{i=0}^{k-1}2^i(n-i)\\ &\,\vdots\\ &=2^nT(0)+\sum_{i=0}^{n-1}2^i(n-i)\\ &=2^n+n(2^n-1)-\sum_{i=0}^{n-1}i2^i\\ &=2^n+n(2^n-1)-\sum_{i=1}^{n-1}\sum_{j=1}^i2^i\\ &=2^n+n(2^n-1)-\sum_{j=1}^{n-1}\sum_{i=j}^{n-1}2^i\\ &=2^n+n(2^n-1)-\sum_{j=1}^{n-1}(2^n-2^j)\\ &=2^n+n(2^n-1)-n2^n+\sum_{j=1}^{n-1}2^j\\ &=2^n-n+2^n-2\\ &=2^{n+1}-n-2\;. \end{align*}$$
This is $\Theta(2^n)$, but it’s not $\Theta(n2^n)$, because it’s not $\Omega(n2^n)$: $$\lim_{n\to\infty}\frac{2^{n+1}-n-2}{n2^n}=0\;.$$