How to calculate the bias of cryptographic hash output from biased input? Assume that I have a hardware entropy source.  Not a TRNG, but the source that might feed a TRNG's extraction algorithm.  This source produces independent bits with a not unreasonable bias of $60/40$.  So that's $p(1) = 0.6$ and $p(0) = 0.4$ and min. entropy would therefore be 0.737 bits/bit.

I then use a cryptographic hash function to extract entropy. Let's say I take groups of 320 bits from my entropy source and hash them with SHA1. What is the resulting bias in the long run output? Or how might this be calculated?

I'm hoping for a numeric answer or calculation rather than verbiage.
Let's call the output of your TNRG $X$—this is a random variable taking the values 0 and 1, with $\Pr[X = 0] = 0.4$ and $\Pr[X = 1] = 0.6$, so that $H_\infty[X] = -\log_2 0.6 \approx 0.737$ bits.  This induces the random variable $Y = \operatorname{SHA-256}(X)$, which has

\begin{align*}
  \Pr[Y = \operatorname{SHA-256}(0)] &amp;= 0.4, \\
  \Pr[Y = \operatorname{SHA-256}(1)] &amp;= 0.6.
\end{align*}

Since $\operatorname{SHA-256}(0) \ne \operatorname{SHA-256}(1)$, we clearly have $H_\infty[Y] = H_\infty[X]$.

What if we have a set of observations $\{X_i\}_i$, which are independent and identically distributed?  Then, e.g.,

\begin{align*}
  \Pr[X_0 = 0, X_1 = 0] &amp;= 0.4^2 &amp;&amp;= 0.16, \\
  \Pr[X_0 = 0, X_1 = 1] &amp;= 0.4 \times 0.6 &amp;&amp;= 0.24, \\
  \Pr[X_0 = 1, X_1 = 0] &amp;= 0.6 \times 0.4 &amp;&amp;= 0.24, \\
  \Pr[X_0 = 1, X_1 = 1] &amp;= 0.6^2 &amp;&amp;= 0.36.
\end{align*}

I leave computing the entropy of the tuple $(X_0, X_1)$ as an exercise for the reader.  What about $\operatorname{SHA-256}(X_0 \mathbin\Vert X_1)$?  Since SHA-256 has no collisions on $00$, $01$, $10$, $11$, once again the entropy is preserved.

In fact, the entropy is not preserved only if there is a collision in the possible inputs.  Obviously the entropy is at most 256 bits for the output of SHA-256 on a single input.  How many distinct outputs are there?  Nobody knows for SHA-256, but for a uniform random function $F\colon \{0,1\}^{256} \to \{0,1\}^{256}$, the expected fraction of distinct outputs is about half the total output space—specifically, $1 - e^{-1}$.  So you might estimate a maximum min-entropy of about 255 bits, in the best case of a perfectly uniform input.

However, you're not limited to feeding in 256 bits; you could feed in 512.  If you feed in $256 + k$ bits to a uniform random function 256-bit function, the expected fraction of distinct outputs is about $1 - e^{-2^k}$ which very rapidly approaches 1 as $k$ grows.  The min-entropy of $(X_0, X_1, \dots, X_{511})$ is obviously $-\log_2 0.6^{512} \approx 377 \gg 256$ bits.  While passing 512 bits through SHA-256 may not necessarily cover the entire 256-bit output space and may not attain the full possible entropy of 256 bits, it's probably a pretty safe bet that the entropy is close enough to 256 bits that the difference doesn't matter to you.  This is plenty to use as a key for a PRNG for cryptographic work.