9 V Electric Lighter
Every real-world source of electrical power has a non-zero internal resistance:







simulate this circuit &ndash; Schematic created using CircuitLab

For typical 9V alkaline batteries, an internal resistance of about 2 Ω is often assumed.

Now, as @JRE said, your 1 cm of (cold) Nichrome wire should have a resistance of ca 0.3 Ω.

Hence, the internal and the load resistance form a voltage divider, with the Voltage across the load (==heating) resistance being

$$
\begin{align}
U_\text{load}&amp;= U_0 \cdot \frac{R_\text{load}}{R_\text{i} + R_\text{load}}\\
&amp;= 9 \text{ V} \cdot \frac{0.3}{2.3}\\
&amp;\approx 1.17 \text{ V}\,\text{.}
\end{align}
$$

Now, knowing that the power lost is 

$$
\begin{align}
P_\text{load}&amp;= U_\text{load} \cdot I_\text{load}\\
&amp;= U_\text{load} \cdot \frac{U_\text{load}}{R_\text{load}}\\
&amp;=\frac{U_\text{load}^2}{R_\text{load}}\\
&amp;\approx\frac{{1.15 \text{ V}}^2}{0.3\,\Omega}\\
&amp;\approx 4.4 \text{ W}\,\text{,}
\end{align}
$$

which isn't very much, but should still feel relatively warm. My guess is that the contact you've made between wire and battery terminals is pretty bad and you're incurring another couple of Ohms resistance here, making the power matching even worse.