Can we treat $i$ as a constant in the inner $j$ summation?
Yes. They are independent of each other. Similar to when you take the derivative of $j \cdot i^{2}$ with respect to $i$ ($\frac{\operatorname{d}}{\operatorname{d}i} j \cdot i^{2} = j \cdot 2 \cdot i$): You treat $j$ like a constant.
You can easily see this by at first taking the inner series and then the outer one:
\begin{align*}
\sum\limits_{i = 0}^{\infty}\left[ \sum\limits_{j = 1}^{n}\left[ \left( i + j \right) \cdot x^{i} \right] \right] &= \sum\limits_{i=0}^{\infty}\left[ \left( i + 1 \right) \cdot x^{i} + \left( i + 2 \right) \cdot x^{i} + \ldots + \left( i + n \right) \cdot x^{i} \right]\\
\end{align*}
So we trat $i$ like a constant (inside the inner sum).
What is the correct approach and steps to solve this?
You can do multiple things. E.g.: You could rewrite the sums like this:
\begin{align*}
\sum\limits_{i = 0}^{\infty}\left[ \sum\limits_{j = 1}^{n}\left[ \left( i + j \right) \cdot x^{i} \right] \right] &= \sum\limits_{j = 1}^{n}\left[ \sum\limits_{i = 0}^{\infty}\left[ \left( i + j \right) \cdot x^{i} \right] \right] = \sum\limits_{j = 1}^{n}\left[ \sum\limits_{i = 0}^{\infty}\left[ i \cdot x^{i} \right] + \sum\limits_{i = 0}^{\infty}\left[ j \cdot x^{i} \right] \right]\\
&= \sum\limits_{j = 1}^{n}\left[ \sum\limits_{i = 0}^{\infty}\left[ i \cdot x^{i} \right] + j \cdot \sum\limits_{i = 0}^{\infty}\left[
x^{i} \right] \right] = \sum\limits_{j = 1}^{n}\left[ \sum\limits_{i = 0}^{\infty}\left[ i \cdot x^{i} \right] \right] + \sum\limits_{j = 1}^{n}\left[ j \cdot \sum\limits_{i = 0}^{\infty}\left[ x^{i} \right] \right]\\
&= \sum\limits_{i = 0}^{\infty}\left[ i \cdot x^{i} \right] \cdot \sum\limits_{j = 1}^{n}\left[ 1 \right] + \sum\limits_{j = 1}^{n}\left[ j \right] \cdot \sum\limits_{i = 0}^{\infty}\left[ x^{i} \right]\\
\end{align*}
We know $\sum\limits_{j = 1}^{n}\left[ 1 \right] = n$ (trivial), $\sum\limits_{j = 1}^{n}\left[ j \right] = \frac{n^{2} + n}{2}$ (this is called a triangular number: see the proof here), $\sum\limits_{i = 0}^{\infty}\left[ x^{i} \right] = \frac{1}{1 - x} \wedge \left| x \right| < 1$ (this is called a infinite geometric series: see a proof here or here) and $\sum\limits_{i = 0}^{\infty}\left[ i \cdot x^{i} \right] = \frac{x}{\left( x - 1 \right)^{2}} \wedge \left| x \right| < 1$ (this is easy to derive via: $\sum\limits_{i = 0}^{\infty}\left[ i \cdot x^{i} \right] = x \cdot \sum\limits_{i = 0}^{\infty}\left[ i \cdot x^{i - 1} \right] = x \cdot \frac{\operatorname{d}\sum\limits_{i = 0}^{\infty}\left[
x^{i} \right]}{\operatorname{d}x} = \frac{\operatorname{d}\frac{1}{1 - x}}{\operatorname{d}x} = \frac{x}{\left( x - 1 \right)^{2}} \wedge \left| x \right| < 1$ - see here). Using this gives us:
\begin{align*}
\sum\limits_{i = 0}^{\infty}\left[ \sum\limits_{j = 1}^{n}\left[ \left( i + j \right) \cdot x^{i} \right] \right] &= n \cdot \frac{x}{\left( x - 1 \right)^{2}} + \frac{n^{2} + n}{2} \cdot \frac{1}{1 - x} = \frac{n \cdot x}{\left( x - 1 \right)^{2}} + \frac{n^{2} + n}{2 \cdot \left( 1 - x \right)}\\
&= \frac{n \cdot x}{\left( x - 1 \right)^{2}} - \frac{n^{2} + n}{2 \cdot \left( x - 1 \right)} = \frac{2 \cdot n \cdot x}{2 \cdot \left( x - 1 \right)^{2}} - \frac{\left( n^{2} + n \right) \cdot \left( x - 1 \right)}{2 \cdot \left( x - 1 \right)^{2}}\\
&= \frac{2 \cdot n \cdot x - \left( n^{2} + n \right) \cdot \left( x - 1 \right)}{2 \cdot \left( x - 1 \right)^{2}} = \frac{-n^{2} \cdot x + n^{2} + n \cdot x + n}{2 \cdot \left( x - 1 \right)^{2}} \wedge \left| x \right| < 1\\
\end{align*}
So:
$$\fbox{$\sum\limits_{i = 0}^{\infty}\left[ \sum\limits_{j = 1}^{n}\left[ \left( i + j \right) \cdot x^{i} \right] \right] = \frac{-n^{2} \cdot x + n^{2} + n \cdot x + n}{2 \cdot \left( x - 1 \right)^{2}} \wedge \left| x \right| < 1$}$$
