If randomly picked from m symbols, each symbol being equally likely, how long on average...

Step 1
The first pick from the box of 26 is guaranteed to produce a letter that you haven't drawn already. To get the second such letter requires on average $\frac{26}{25}$ picks — close to one more, but a little bit more than that because you have a small chance of drawing the same letter as you did the first time. To get the third distinct letter requires on average $\frac{26}{24}$ picks, and so on.
To pick at least k different elements from a box containing n different elements one expects to need a total of
1$1+\frac{n}{n-1}+\frac{n}{n-2}+\dots +\frac{n}{n-k+1}=n\sum _{i=1}^k\frac{1}{n-i+1}$ picks.
Step 2
For your example of $n=26,k=13$, this is approximately 17.5 picks.
You wanted to know what happens when $k=\frac{n}{2}$. When n is not too small the number of picks required to get about $\frac{n}{2}$ different items can be shown to be approximately
0.69n
where $0.69\approx \log 2$. (To show this, write the formula above as
$n\sum _{i=1}^k\frac{1}{n-i+1}=n(H_n-H_{n-k})$
where $H_i$ are the harmonic numbers, and use the approximation $H_n\approx \log n+\gamma$)