1. **Base Case for $n = 1$:**
- For $n = 1$, the condition is $\tan x_1 = 2^{\frac{1}{2}} = \sqrt{2}$.
- We need to show $\cos x_1 \le \frac{1}{\sqrt{3}}$.
- Since $\tan x_1 = \sqrt{2}$, we have $\sin x_1 = \frac{\sqrt{2}}{\sqrt{3}}$ and $\cos x_1 = \frac{1}{\sqrt{3}}$.
- Therefore, $\cos x_1 = \frac{1}{\sqrt{3}} \le \frac{1}{\sqrt{3}}$, which is true.

2. **Case for $n = 2$:**
- Given $\tan x_1 \tan x_2 = 2$.
- We need to show $\cos x_1 + \cos x_2 \le \frac{2}{\sqrt{3}}$.
- Let $\alpha = x_1$ and $\beta = x_2$. We need to show that if $\cos \alpha + \cos \beta > \frac{2}{\sqrt{3}}$, then $\tan \alpha \tan \beta < 2$.
- Equivalently, if $a = \cos \alpha$ and $b = \cos \beta$ with $a + b > \frac{2}{\sqrt{3}}$, we need to show $\left(\frac{1}{a^2} - 1\right)\left(\frac{1}{b^2} - 1\right) < 2$.
- To show this, consider $a + b = \frac{2}{\sqrt{3}} + \epsilon$ for small $\epsilon$. Then $b = \frac{2}{\sqrt{3}} + \epsilon - a$.
- We need to show $\left(\frac{1}{a^2} - 1\right)\left(\frac{1}{(\frac{2}{\sqrt{3}} + \epsilon - a)^2} - 1\right) \le 2$.
- This can be verified by taking derivatives and checking the maximum value, which is left as an exercise.

3. **Case for $n = 3$:**
- Given $\tan x_1 \tan x_2 \tan x_3 = 2^{\frac{3}{2}} = 2\sqrt{2}$.
- We need to show $\cos x_1 + \cos x_2 + \cos x_3 \le 2$.
- Let $\alpha = x_1$, $\beta = x_2$, and $\gamma = x_3$. We need to show that if $\cos \alpha + \cos \beta + \cos \gamma > 2$, then $\tan \alpha \tan \beta \tan \gamma < 2\sqrt{2}$.
- Equivalently, if $a = \cos \alpha$, $b = \cos \beta$, and $c = \cos \gamma$ with $a + b + c = 2$, we need to show $\left(\frac{1}{a^2} - 1\right)\left(\frac{1}{b^2} - 1\right)\left(\frac{1}{c^2} - 1\right) \le 2$.
- By symmetry and the properties of logarithms, we can assume $a = b$. Then we need to show $\left(\frac{1}{a^2} - 1\right)^2\left(\frac{1}{(2 - 2a)^2} - 1\right) \le 2$ for $a \in (\frac{1}{2}, 1)$.
- This can be verified by taking derivatives and checking the maximum value, which is left as an exercise.

4. **Induction for $n > 3$:**
- Assume the result holds for $n = k$. We need to show it holds for $n = k + 1$.
- Given $\tan x_1 \tan x_2 \cdots \tan x_{k+1} = 2^{\frac{k+1}{2}}$.
- There must be an $x_i$ with $\tan x_i \le \sqrt{2}$. Remove this $x_i$, and the product of the remaining $x_j$'s is at least $2^{\frac{k}{2}}$.
- By the inductive hypothesis, the sum of the cosines of the remaining $x_j$'s is at most $k - 1$.
- Noting that $\cos x_i \le 1$, we have $\sum_{i=1}^{k+1} \cos x_i \le (k - 1) + 1 = k$.

5. Optimality:
- For $n \ge 3$, let $x_1, x_2, \ldots, x_{n-1}$ be $\epsilon$ for small $\epsilon$ and define $x_n$ accordingly.
- For $n = 1, 2$, let all $x_i$'s be equal.

Thus, the least $\lambda$ is $\frac{n}{\sqrt{3}}$ for $n = 1, 2$ and $n - 1$ for $n \ge 3$.

The final answer is $\boxed{ \lambda = \frac{n}{\sqrt{3}} }$ for $n = 1, 2$ and $\lambda = n - 1$ for $n \ge 3$.