Asymptotic variance of a single parameter in linear regression Consider the linear regression model
$$ Y_i = \beta_0 + \beta_1 x_{i1} + \beta_2 x_{i2} + \dots + \beta_k x_{ik} + \epsilon_i $$
or equivalently in matrix norm
$$ \mathbf{Y} = \beta \mathbf{X} + \epsilon$$
where $\mathbb{E}[\epsilon|X] = 0$, $\mathbf{X}$ is $n \times (k + 1)$ matrix. It can be shown by the Central Limit Theorem that the OLS estimates satisfy
$$ \sqrt{n}(\hat{\beta} - \beta) \to^d \mathcal{N}(0, \Sigma) $$
where $\Sigma = \mathbb{E}[XX']^{-1} \mathbb{E}[XX'\epsilon^2] \mathbb{E}[XX']^{-1}$ is a $(k+1) \times (k+1)$ matrix.

Suppose we were interested in inference only on $\beta_1$. Is there a way to analytically derive the asymptotic variance of only $\hat{\beta}_1$? More precisely, the expression for $\sigma^2_{\beta_1}$ that satisfies
$$ \sqrt{n}(\hat{\beta}_1 - \beta_1) \to^d \mathcal{N}(0, \sigma^2_{\beta_1}) $$
I think it should be the $(2,2)$ entry of $\Sigma$, though I'm not sure how to separate it out analytically given the matrix inverses. Computationally, we can just take the relevant entry $\hat{\Sigma}_{2,2}$ from the full estimated variance matrix but this would be very inefficient (especially if $k$ large) since we only need a single entry. I was trying some ideas with the Frisch-Waugh-Lovell Theorem but not quite getting anywhere.
Any ideas?

We may "need a single entry" only, but all the sample will participate in computing it.
Write your model as
$$Y_i = \beta_1 x_{i1} + \beta_0 +  + \beta_2 x_{i2} + \dots + \beta_k x_{ik} + \epsilon_i$$
and partition the $n \times k$ regressore matrix as
$$\mathbf X = \left [\mathbf x_1\quad Z \right],$$
where $Z$ contains all other regressors including the constant. Let $D = {\rm diag} \{\hat \epsilon^2_i\}$, a $n \times n$ diagonal matrix. Then, in practice,
$$\widehat \Sigma = n\left [\begin{matrix} \mathbf x_1'\mathbf x_1 & \mathbf x_1'Z \\ Z'\mathbf x_1 & Z'Z \end{matrix}\right]^{-1}\left [\begin{matrix} \mathbf x_1'D\mathbf x_1 & \mathbf x_1'DZ \\ Z'D\mathbf x_1 & Z'DZ \end{matrix}\right]\left [\begin{matrix} \mathbf x_1'\mathbf x_1 & \mathbf x_1'Z \\ Z'\mathbf x_1 & Z'Z \end{matrix}\right]^{-1}.$$
The upper left element will be $1 \times 1$ and it is what you want.
Apply the most convenient matrix blockwise inversion formula, and obtain the final upper left element of $\widehat \Sigma$. Determine whether computing only it results in improvements as regards computational efficiency.
NOTE 1: You need to correct the expression for $\Sigma$ in your post (the expected value should be under the inverse sign).
NOTE 2: The $n$ in front of the expression for $\widehat \Sigma$ is because we compute the variance of $\sqrt{n}(\hat{\beta} - \beta)$. If we want the approximation for the finite sample variance of $\hat \beta$, we ignore it.
