Asymptotics of a logarithmic series
First, the change-of-base formula may be helpful here:
$$
\log_b(x) = \frac{\ln x}{\ln b}
$$
Also, with sums I always find it helps to write out the sum explicitly, so let's do that:
$$
T(n) = \sum_{i=2}^{n} \log_i n
= \log_2(n) + \log_3(n) + \log_4(n) + \cdots + \log_n(n)
$$

Now if you want to show this is $\theta(n)$, there are two things: you should show it is lower bounded by $cn$, and it is upper bounded by $Cn$, for some constants $c$ and $C$.


For the lower bound, use the fact that each term is greater than or equal to $1$.
For the upper bound, I suggest dividing the sum into two parts: from $i = 2$ to $i = \sqrt{n}$, and $i = \sqrt{n}$ to $n$. So the first part is
$$
\left(\log_2(n) + \log_3(n) + \log_4(n) + \cdots + \log_{\lfloor\sqrt{n}\rfloor}(n)\right).
$$
Here we have $\sqrt{n}$ terms, each which is at most $\log_2(n)$. The second part is
$$
\left( \log_{\lceil \sqrt{n} \rceil}(n) + \cdots + \log_n(n) \right).
$$
Here we have at most $n$ terms, but you can show (using the change of base formula above) that each term is at most $2$.


Now adding up the two parts, you should get an upper bound of
$$
\sqrt{n} \log_2(n) + 2n,
$$
so the final step is to show this is $O(n)$.