If I understand correctly, you are interested in the expected value of the magnitude of the win (or) loss (and not the magnitude of the expected value of the win (or) loss). Hence, you are interested in computing $\mathbb{E}(S_n)$ of the underlying random variable where $S_n = \left| X_1 + X_2 + \cdots + X_n \right|.$ Note that $S_n$ can take values in the set $\{0,5,10,\ldots,5n\}$.
Now let us find out the probability that $S_n = 5m$ for some $m \in \{0,1,2,\ldots,n\}$.
This means we are interested in the event $X_1 + X_2 + \cdots + X_n = \pm 5m$.
Let us first evaluate the probability of the event $X_1 + X_2 + \cdots + X_n = 5m$.
For this event to occur, if you lose $k$ times, you need to win $m+k$ times and get back your stake the remaining $n-m-2k$ times where $k \in \left \{0,1,2,\ldots, \left \lfloor \frac{n-m}{2} \right \rfloor \right\}$.
Hence, the desired probability of the event $X_1 + X_2 + \cdots + X_n = 5m$ is given by $\sum_{k=0}^{\left \lfloor \frac{n-m}{2} \right \rfloor} \frac{n!}{k!(m+k)!(n-m-2k)!} \frac1{3^n}$
Hence, the desired probability of the event $S_n = 5m$ is given by $P_n(m) = \sum_{k=0}^{\left \lfloor \frac{n-m}{2} \right \rfloor} \frac{n!}{k!(m+k)!(n-m-2k)!} \frac2{3^n}$
Hence, the expected value of $S_n$ is given by $\sum_{m=0}^{n} 5m P_n(m) = \sum_{m=0}^{n} \sum_{k=0}^{\left \lfloor \frac{n-m}{2} \right \rfloor} \frac{n!}{k!(m+k)!(n-m-2k)!} \frac{10m}{3^n}$