Incredible Blackjack Hand

We can assume that the dealer's hole card is irrelevant. So we are looking for the probability that the first six cards are $8$ out of the $415$ cards left in the deck (excluding the dealer's up-card $9$). Among the $415$ cards, there are $32$ $8$'s. There are $\binom{32}{6}$ ways of choosing six $8$'s, and there are a total of $\binom{415}{6}$ ways of choosing the first six cards. Hence, the probability is $$\frac{\binom{32}{6}}{\binom{415}{6}} \approx 1.32\times 10^{-7},$$ which is very very small.

In 8 decks there are 32 8s. To pick up six of them there are $\binom{32}{6}=906,192$ possible ways for that to happen. There are, similarly, $\binom{416}{6}=6,942,219,827,088$ ways to get just any six cards. Dividing these, that's about a 1 in 7.6 million chance for this to happen. This is about 1/12 as likely as drawing a royal flush on five cards in a single deck.

Of course, you asked about six of any card, which is considerably easier (13 times more common than above), somewhat more likely than a royal flush.

$8$ decks gives a total of $52 \times 8 = 416$ cards.

$8$ decks with 4 cards each as an Eight gives a total of $32$ possible Eights to draw.

So a probability of drawing each Eight in sequence is:

$$\underbrace{\frac{32}{416} \times \frac{31}{415} \times \dots \frac{28}{412} \times \frac{27}{411}}_{\text{6 draws}}$$

If you take into account that the dealer doesn't draw an Eight, then you have $415$ cards to choose from, so a more accurate probability is:

$$\frac{32}{415} \times \frac{31}{414} \times \dots \frac{28}{411} \times \frac{27}{410}$$

Since there are 13 possible cards that can be drawn in sequence, the answer to

what the odds are in getting 6 straight cards of the same denomination from an 8 deck shoe

is

$$13 \times \frac{32}{415} \times \frac{31}{414} \times \dots \frac{28}{411} \times \frac{27}{410}$$

which is about $1$ in every $580,798$ attempts. Keep in mind that if you split all the Tens out of the deck, any other competent player will be angry and leave the table.

Clarification: I have interpreted the phrase "what are the odds in getting 6 straight cards of the same denomination from an 8 deck shoe" as asking for the chance that there are 6 straight cards of the same denomination from an 8 deck shoe. My answer addresses this interpretation only.

Ignoring the details of the game, I'll just consider the chance that a well-shuffled 8-deck shoe has 6 or more eights in a row somewhere. I solved a similar problem here.

For your problem, put $b=384$ and $w=32$ in my answer above to arrive at

\begin{eqnarray*} \mathbb{P}(\mbox{at least 6 eights in a row}) &=&{378917534435104330038751954618647 \over 7539892080833060495675366062952229323}\\[5pt] &=&0.000050255,\end{eqnarray*} or about 1 in 20,000.

If you ask for the probability of at least 6 in a row of any of the 13 possible values, an approximate answer is to multiply the above by 13, giving $P\approx .0006533$ or about 1 in 1350. Not a common occurrence, but not that rare!