Why decision tree method for lower bound on finding a minimum doesn't work
In the comments I said that $n$ is the number of different possible outcomes of a minimum algorithm and that this is only a lower bound for the number of leaves of the decision trees of the comparison-based minimum algorithms (in which nodes represent the comparisons and edges the results of those comparisons). This is an application of Lemma 1 here, which then implies the bound that you got on the height of the tree. This is Lemma 2 in the linked document. They call this an information theoretic argument.
The reason why it gives such a conservative lower bound for the height of the trees, is that the bound on the number of leaves is too conservative. So, Lemma 1's fault of not being strong enough.

I was curious to see if this method could give a better lower bound if one manages to bound better the number of leaves. For this what we need to do is to have some control on the number of times an outcome of the algorithm is repeated across the leaves.
Suppose a comparison-based algorithm for minimum has decision tree with a maximum of $K$ leaves corresponding to the same outcome.
Then the tree has at most $Kn$ leaves and at most height $\log(n)+\log(K)$. If $n&gt;2(\log(n)+\log(K))$, then none of the paths from root to leaf compares all elements of the array. Since the height of the path is bounded by $\log(n)+\log(K)$, then there are at most that many comparisons in the path. Each comparison looks at $2$ elements, so at most $2(\log(n)+\log(K))$ are inspected. Then we can keep the elements compared in one of the paths fixed and make the other elements smaller than the element that that path chooses as minimum. This input would result on the same output from the algorithm but have a different minimum. So, the algorithm must be incorrect. So, the inequality above must be false. This argument has the flavor of the adversary argument, which applied directly gives the best lower bound for this problem.
Hence $n\leq 2(\log(n)+\log(K))$. From this we get that $\log(K)\geq n/2-\log(n)$, or in other words
$$K\geq 2^{n/2-\log(n)}$$
Now, this tells us that the decision trees for these algorithms will always have at least $K+(n-1)\geq 2^{n/2-\log(n)}+(n-1)$ leaves. So, their height is at least
$$\log\left(2^{n/2-\log(n)}+(n-1)\right)\geq n/2-\log(n)$$