OK, I've been very confused by this notation, but are you sure that the answer is $\Theta(n\cdot2^n)$?
Your tree analysis is correct: at the first step, you add $\Theta(n)$, at the next one, you add $2 \Theta(n-1)$, at the next one $4\Theta(n-2)$, etc. In other words, at the $i$-th step, you add $2^i \Theta(n-i)$. Because you need to add all the steps, the actual sum is then $T(n) = \sum_{i=0}^n 2^i \Theta(n-i)$ Now, since $\Theta(n)$ means "a function doing $n$ steps of computation", you can rewrite $\Theta(n-i)$ as $\Theta(n)-\Theta(i)$. The sum becomes $ T(n) = \sum_{i=0}^n 2^i \Theta(n) - 2^i\Theta(i) = \sum_{i=0}^n 2^i \Theta(n) - \sum_{i=0}^n 2^i\Theta(i) $ If we pull $\Theta(n)$ out of the sum, we can use the fact that $\sum_{i=0}^n 2^i = 2^{n+1} - 1$. Moreover, we can use the fact that $\sum_{i=0}^n i\cdot 2^i = 2(n\cdot 2^n - 2^n+1)$. The sum then becomes: $ T(n) = n*(2^{n+1} - 1) - 2(n\cdot 2^n - 2^n + 1) = -n + 2^{n+1} - 2 $ So I would conclude that $T(n)= \Theta(2^{n})$.
Actually, according to Wolfram Alpha, we have $t(n) = -n + 3\cdot2^n - 2$.
So either I also made a mistake in the calculation (which is obviously possible), or the answer is indeed $\Theta(2^n)$, as you found yourself.