You cannot meaningfully forecast this: it is a random lottery draw. The best you can do is work out the probability of a particular sum.
As far as I can tell it is the sum of $7$ numbers sampled without replacement from $1$ to $45$. The minimum value is $28$ and the largest $294$. Ignoring the order in which the balls come out there are ${45 \choose 7}=45379620 $ possibilities (or $\frac{45!}{38!}=228713284800$ if you do take order into account).
As for the number of ways of getting a particular sum $s$, this is the number of partitions of $s$ into $7$ distinct parts each no more than $45$. You can use generating functions or recurrences to find this, but I already have a java applet to do this. For instance for the most likely outcome $s=161$ which turned up on 21 May 2012, choose
- "Partitions with distinct terms of:" 161
- "Exact number of terms:" 7
- "Each term no more than:" 45
then click on "calculate partitions" to get the answer $554256$ and so a probability of $\frac{554256}{45379620} \approx 0.0122$.
The equivalent for a sum of $81$ as in 6 February was $\frac{23573}{45379620} \approx 0.0005$.